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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% The Project Gutenberg EBook of The Integration of Functions of a Single %
+% Variable, by G. H. Hardy                                                %
+%                                                                         %
+% This eBook is for the use of anyone anywhere at no cost and with        %
+% almost no restrictions whatsoever.  You may copy it, give it away or    %
+% re-use it under the terms of the Project Gutenberg License included     %
+% with this eBook or online at www.gutenberg.org                          %
+%                                                                         %
+%                                                                         %
+% Title: The Integration of Functions of a Single Variable                %
+%                                                                         %
+% Author: G. H. Hardy                                                     %
+%                                                                         %
+% Editor: P. Hall                                                         %
+%         F. Smithies                                                     %
+%                                                                         %
+% Release Date: September 23, 2012 [EBook #38993]                         %
+%                                                                         %
+% Language: English                                                       %
+%                                                                         %
+% Character set encoding: ISO-8859-1                                      %
+%                                                                         %
+% *** START OF THIS PROJECT GUTENBERG EBOOK INTEGRATION OF FUNCTIONS OF ONE VARIABLE ***
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+%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\begin{document}
+\PGBoilerPlate
+{\small
+\begin{PGtext}
+The Project Gutenberg EBook of The Integration of Functions of a Single
+Variable, by G. H. Hardy
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+
+Title: The Integration of Functions of a Single Variable
+
+Author: G. H. Hardy
+
+Editor: P. Hall
+        F. Smithies
+
+Release Date: September 23, 2012 [EBook #38993]
+
+Language: English
+
+Character set encoding: ISO-8859-1
+
+*** START OF THIS PROJECT GUTENBERG EBOOK INTEGRATION OF FUNCTIONS OF ONE VARIABLE ***
+\end{PGtext}
+\clearpage
+\begin{PGtext}
+Produced by Brenda Lewis, Anna Hall and the Online
+Distributed Proofreading Team at http://www.pgdp.net (This
+file was produced from images generously made available
+by The Internet Archive/American Libraries.)
+\end{PGtext}}
+\vfill
+
+\BookMark{0}{Transcriber's Note}
+\subsection*{\centering\normalfont\scshape
+\normalsize\MakeLowercase{\TransNote}}
+
+\TransNoteText
+\clearpage
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\FrontMatter
+
+%% -----File: 001.png---
+\begin{center}
+{\large Cambridge Tracts in Mathematics\\
+and Mathematical Physics\\}
+
+\smallskip
+
+\textsc{General Editors\\
+P.~HALL, F.R.S. and F.~SMITHIES, Ph.D.}
+
+\bigskip
+
+No.~2
+
+\smallskip
+
+{\LARGE THE\\
+INTEGRATION OF FUNCTIONS\\
+OF A SINGLE VARIABLE\\}
+
+\smallskip
+
+{\large BY}
+
+{\Large G.~H. HARDY}
+\vfill
+{\includegraphics[width=1in]{./images/device.pdf}} %Publisher's device
+
+{\large CAMBRIDGE UNIVERSITY PRESS\\}
+%% -----File: 002.png---
+%[Blank Page]
+%% -----File: 003.png---
+\newpage
+{\Large Cambridge Tracts in Mathematics\\
+and Mathematical Physics\\}
+
+\medskip
+
+\textsc{General Editors\\
+P.~HALL, F.R.S. and F.~SMITHIES, Ph.D.}
+
+\vfill
+
+{\large No.~2}
+
+\bigskip
+
+{\large The Integration of Functions of a\\
+Single Variable}
+
+\vfill
+%% -----File: 004.png---
+%[Blank Page]
+%% -----File: 005.png---
+\newpage
+{\large THE\\}
+{\LARGE INTEGRATION OF FUNCTIONS\\
+OF A SINGLE VARIABLE\\}
+
+\vfill
+
+BY
+
+\medskip
+
+{\large G. H. HARDY}
+
+\vfill
+
+SECOND EDITION
+
+\vfill
+
+{\large CAMBRIDGE\\
+AT THE UNIVERSITY PRESS\\
+1966\\}
+
+\bigskip
+%% -----File: 006.png---
+\newpage
+PUBLISHED BY
+
+THE SYNDICS OF THE CAMBRIDGE UNIVERSITY PRESS
+
+Bentley House, 200 Euston Road, London, N.W. 1\\
+American Branch: 32 East 57th Street, New York, N.Y. 10022
+
+\vfill
+
+\begin{tabular}{l l}
+  \textit{First Edition} & 1905 \\
+  \textit{Second Edition} & 1916 \\
+  \textit{Reprinted} & 1928 \\
+    & 1958 \\
+    & 1966 \\
+\end{tabular}
+
+\vfill
+
+\textit{First printed in Great Britain at the University Press, Cambridge
+Reprinted by offset-litho by Jarrold \& Sons Ltd., Norwich}
+\end{center}
+%% -----File: 007.png---
+\newpage
+\Section{PREFACE}
+
+This tract has been long out of print, and there is still some
+demand for it. I did not publish a second edition before,
+because I intended to incorporate its contents in a larger treatise on
+the subject which I had arranged to write in collaboration with
+Dr~Bromwich. Four or five years have passed, and it seems very
+doubtful whether either of us will ever find the time to carry out
+our intention. I have therefore decided to republish the tract.
+
+The new edition differs from the first in one important point
+only. In the first edition I reproduced a proof of Abel's which
+Mr~J.~E. Littlewood afterwards discovered to be invalid. The
+correction of this error has led me to rewrite a few sections (\hyperlink{5 para 11.}{pp.~36--41}
+of the present edition) completely. The proof which I give now is
+due to Mr~H.~T.~J. Norton. I am also indebted to Mr~Norton,
+and to Mr~S. Pollard, for many other criticisms of a less important
+character.
+
+\begin{flushright}
+G. H. H.
+\end{flushright}
+
+\textit{January} 1916.
+%% -----File: 008.png---
+%[Blank Page]
+%% -----File: 009.png---
+\newpage
+\Contents{CONTENTS}
+
+\hfill \textsc{page}
+\begin{description}
+\item[I.] Introduction \hfill \pageref{I. Introduction}
+\item[II.] Elementary functions and their classification \hfill \pageref{II. Elementary functions and their classification}
+\item[III.] The integration of elementary functions. Summary of results \hfill \pageref{III. The integration of elementary functions. Summary of results}
+\item[IV.] The integration of rational functions \hfill \pageref{IV. Rational functions}
+\begin{description}
+\item[1--3.] The method of partial fractions \hfill \pageref{4 para 1.}
+\item[4.] Hermite's method of integration \hfill \pageref{4 para 4.}
+\item[5.] Particular problems of integration \hfill \pageref{4 para 5.}
+\item[6.] The limitations of the methods of integration \hfill \pageref{4 para 6.}
+\item[7.] Conclusion \hfill \pageref{4 para 7.}
+\end{description}
+\item[V.] The integration of algebraical functions \hfill \pageref{V. Algebraical Functions}
+\begin{description}
+\item[1.] Algebraical functions \hfill \pageref{5 para 1.}
+\item[2.] Integration by rationalisation. Integrals associated with
+conics \hfill \pageref{5 para 2.}
+\item[3--6.] The integral $\int R\{x, \sqrt{ax^{2} + 2bx + c}\}\,dx$ \hfill \pageref{5 para 3.}
+\item[7.] Unicursal plane curves \hfill \pageref{5 para 7.}
+\item[8.] Particular cases \hfill \pageref{5 para 8.}
+\item[9.] Unicursal curves in space \hfill \pageref{5 para 9.}
+\item[10.] Integrals of algebraical functions in general \hfill \pageref{5 para 10.}
+\item[11--14.] The general form of the integral of an algebraical function.
+Integrals which are themselves algebraical \hfill \pageref{5 para 11.}
+\item[15.] Discussion of a particular case \hfill \pageref{5 para 15.}
+\item[16.] The transcendence of $e^{x}$ and $\log x$ \hfill \pageref{5 para 16.}
+\item[17.] Laplace's principle \hfill \pageref{5 para 17.}
+\item[18.] The general form of the integral of an algebraical function
+(\emph{continued}). Integrals expressible by algebraical functions
+and logarithms \hfill \pageref{5 para 18.}
+%% -----File: 010.png---
+\item[19.] Elliptic and pseudo-elliptic integrals. Binomial integrals \hfill \pageref{5 para 19.}
+\item[20.] Curves of deficiency~$1$. The plane cubic \hfill \pageref{5 para 20.}
+\item[21.] Degenerate Abelian integrals \hfill \pageref{5 para 21.}
+\item[22.] The classification of elliptic integrals \hfill \pageref{5 para 22.}
+\end{description}
+\item[VI.] The  integration of transcendental functions \hfill \pageref{VI. Transcendental functions}
+\begin{description}
+\item[1.] Preliminary \hfill \pageref{6 para 1.}
+\item[2.] The integral $\int R (e^{ax}, e^{bx}, \dots, e^{kx})\,dx$ \hfill \pageref{6 para 2.}
+\item[3.] The integral $\int P(x, e^{ax}, e^{bx}, \dots)\,dx$ \hfill \pageref{6 para 3.}
+\item[4.] The integral $\int e^{x} R(x)\, dx$. The logarithm-integral \hfill \pageref{6 para 5.}
+\item[5.] Liouville's general theorem \hfill \pageref{6 para 5.}
+\item[6.] The integral $\int \log x R(x)\, dx$ \hfill \pageref{6 para 6.}
+\item[7.] Conclusion \hfill \pageref{6 para 7.}
+\end{description}
+\item[Appendix I.] Bibliography \hfill \pageref{Appendix I}
+\item[Appendix II.] On Abel's proof of the theorem of v., §~11 \hfill \pageref{Appendix II}
+\end{description}
+%% -----File: 011.png---
+\MainMatter
+\cleardoublepage
+\Section{THE INTEGRATION OF FUNCTIONS OF
+A SINGLE VARIABLE}
+
+\Section{I. Introduction}
+
+The problem considered in the following pages is what is sometimes
+called the problem of `indefinite integration' or of `finding a function
+whose differential coefficient is a given function'. These descriptions
+are vague and in some ways misleading; and it is necessary to define
+our problem more precisely before we proceed further.
+
+Let us suppose for the moment that $f(x)$~is a real continuous
+function of the real variable~$x$. We wish to determine a function~$y$
+whose differential coefficient is~$f(x)$, or to solve the equation
+\[
+\frac{dy}{dx} = f(x).
+\Tag{(1)}
+\]
+A little reflection shows that this problem may be analysed into a
+number of parts.
+
+We wish, first, to know whether such a function as~$y$ necessarily
+exists, whether the equation~\Eq{(1)} has always a solution; whether the
+solution, if it exists, is unique; and what relations hold between
+different solutions, if there are more than one. The answers to these
+questions are contained in that part of the theory of functions of a
+real variable which deals with `definite integrals'. The definite
+integral
+\[
+y = \int_{a}^{x} f(t)\,dt,
+\Tag{(2)}
+\]
+which is defined as the limit of a certain sum, is a solution of the
+equation~\Eq{(1)}. Further
+\[
+y + C,
+\Tag{(3)}
+\]
+where $C$~is an arbitrary constant, is also a solution, and all solutions of~\Eq{(1)}
+are of the form~\Eq{(3)}.
+%% -----File: 012.png---
+
+These results we shall take for granted. The questions with which
+we shall be concerned are of a quite different character. They are
+questions as to the functional form of~$y$ when $f(x)$~is a function of
+some stated form. It is sometimes said that the problem of indefinite
+integration is that of `finding an actual expression for~$y$ when $f(x)$~is
+given'. This statement is however still lacking in precision. The theory
+of definite integrals provides us not only with a proof of the existence
+of a solution, but also with an expression for it, an expression in the
+form of a limit. The problem of indefinite integration can be stated
+precisely only when we introduce sweeping restrictions as to the classes
+of functions and the modes of expression which we are considering.
+
+Let us suppose that $f(x)$ belongs to some special class of functions~$\FF$.
+Then we may ask whether $y$~is itself a member of~$\FF$, or can be
+expressed, according to some simple standard mode of expression, in
+terms of functions which are members of~$\FF$. To take a trivial
+example, we might suppose that $\FF$~is the class of polynomials with
+rational coefficients: the answer would then be that $y$~is in all cases
+itself a member of~$\FF$.
+
+The range and difficulty of our problem will depend upon our
+choice of (1)~a class of functions and (2)~a standard `mode of expression'.
+We shall, for the purposes of this tract, take $\FF$ to be the
+class of \emph{elementary functions}, a class which will be defined precisely in
+the next section, and our mode of expression to be that of \emph{explicit
+expression in finite terms}, \textit{i.e.}\ by formulae which do not involve passages
+to a limit.
+
+One or two more preliminary remarks are needed. The subject-matter
+of the tract forms a chapter in the `integral calculus'\footnotemark,
+  \footnotetext{Euler, the first systematic writer on the `integral calculus', defined it in
+  a manner which identifies it with the theory of differential equations: `calculus
+  integralis est methodus, ex data differentialium relatione inveniendi relationem
+  ipsarum quantitatum' (\textit{Institutiones calculi integralis}, p.~1). We are concerned
+  only with the special equation~\Eq{(1)}, but all the remarks we have made may be
+  generalised so as to apply to the wider theory.}%
+but
+does not depend in any way on any direct theory of integration. Such
+an equation as
+\[
+y = \int f(x)\, dx
+\Tag{(4)}
+\]
+is to be regarded as merely another way of writing~\Eq{(1)}: the integral
+sign is used merely on grounds of technical convenience, and might
+be eliminated throughout without any substantial change in the
+argument.
+%% -----File: 013.png---
+
+The variable~$x$ is in general supposed to be complex. But the tract
+should be intelligible to a reader who is not acquainted with the theory
+of analytic functions and who regards~$x$ as real and the functions of~$x$
+which occur as real or complex functions of a real variable.
+
+The functions with which we shall be dealing will always be such
+as are regular except for certain special values of~$x$. These values of~$x$
+we shall simply ignore. The meaning of such an equation as
+\[
+\int \frac{dx}{x} = \log x
+\]
+is in no way affected by the fact that $1/x$~and~$\log x$ have infinities for
+$x = 0$.
+
+\Section{II. Elementary functions and their classification}
+
+An \emph{elementary function} is a member of the class of functions which
+comprises
+
+\Item{(i)} rational functions,
+
+\Item{(ii)} algebraical functions, explicit or implicit,
+
+\Item{(iii)} the exponential function~$e^{x}$,
+
+\Item{(iv)} the logarithmic function~$\log x$,
+
+\Item{(v)} all functions which can be defined by means of any finite
+combination of the symbols proper to the preceding four classes of
+functions.
+
+A few remarks and examples may help to elucidate this definition.
+
+\Paragraph{2}{1.} A \emph{rational function} is a function defined by means of any finite
+combination of the elementary operations of addition, multiplication,
+and division, operating on the variable~$x$.
+
+It is shown in elementary algebra that any rational function of~$x$
+may be expressed in the form
+\[
+f(x) = \frac{a_{0}x^{m} + a_{1}x^{m-1} + \dots + a_{m}}
+            {b_{0}x^{n} + b_{1}x^{n-1} + \dots + b_{n}},
+\]
+where $m$~and~$n$ are positive integers, the $a$'s~and~$b$'s are constants, and
+the numerator and denominator have no common factor. We shall
+adopt this expression as the standard form of a rational function. It
+is hardly necessary to remark that it is in no way involved in the
+%% -----File: 014.png---
+definition of a rational function that these constants should be rational
+or algebraical\footnote
+  {An algebraical number is a number which is the root of an algebraical equation
+  whose coefficients are integral. It is known that there are numbers (such as
+  $e$~and~$\pi$) which are not roots of any such equation. See, for example, Hobson's
+  \textit{Squaring the circle} (Cambridge, 1913).}
+or real \emph{numbers}. Thus
+\[
+\frac{x^{2} + x + i \sqrt{2}}{x\sqrt{2} - e}
+\]
+is a rational function.
+
+\Paragraph{2}{2.} An \emph{explicit algebraical function} is a function defined by means
+of any finite combination of the four elementary operations and any
+finite number of operations of root extraction. Thus
+\[
+\frac{\sqrt{1+x} - \sqrt[3]{1-x}}
+     {\sqrt{1+x} + \sqrt[3]{1-x}},\quad
+\sqrt{x +\sqrt{x + \sqrt{x}}},\quad
+\left(\frac{x^{2} + x + i \sqrt{2}}{x\sqrt{2} - e}\right)^{\frac{2}{3}}
+\]
+are explicit algebraical functions. And so is~$x^{m/n}$ (\textit{i.e.}~$\sqrt[n]{x^{m}}$) for any
+integral values of $m$~and~$n$. On the other hand
+\[
+x^{\sqrt{2}},\quad x^{1 + i}
+\]
+are not algebraical functions at all, but transcendental functions, as
+irrational or complex powers are defined by the aid of exponentials
+and logarithms.
+
+Any explicit algebraical function of~$x$ satisfies an equation
+\[
+P_{0}y^{n} + P_{1}y^{n-1} + \dots + P_{n} = 0
+\]
+whose coefficients are polynomials in~$x$. Thus, for example, the
+function
+\[
+y = \sqrt{x} +\sqrt{x +\sqrt{x}}
+\]
+satisfies the equation
+\[
+y^{4} - (4y^{2} + 4y + 1)x = 0.
+\]
+The converse is not true, since it has been proved that in general
+equations of degree higher than the fourth have no roots which are
+explicit algebraical functions of their coefficients. A simple example
+is given by the equation
+\[
+y^{5} - y - x = 0.
+\]
+We are thus led to consider a more general class of functions, \emph{implicit}
+algebraical functions, which includes the class of explicit algebraical
+functions.
+%% -----File: 015.png---
+
+\Paragraph{2}{3.} An \emph{algebraical function} of~$x$ is a function which satisfies an
+equation
+\[
+P_{0}y^{n} + P_{1}y^{n-1} + \dots + P_{n} = 0
+\Tag{(1)}
+\]
+whose coefficients are polynomials in~$x$.
+
+Let us denote by~$P(x, y)$ a polynomial such as occurs on the left-hand
+side of~\Eq{(1)}. Then there are two possibilities as regards any
+particular polynomial~$P(x, y)$. Either it is possible to express~$P(x, y)$
+as the product of two polynomials of the same type, neither of which
+is a mere constant, or it is not. In the first case $P(x, y)$~is said to
+be \emph{reducible}, in the second \emph{irreducible}. Thus
+\[
+y^{4} - x^{2} = (y^{2} + x)(y^{2} - x)
+\]
+is reducible, while both $y^{2} + x$ and $y^{2} - x$ are irreducible.
+
+The equation~\Eq{(1)} is said to be reducible or irreducible according as
+its left-hand side is reducible or irreducible. A reducible equation can
+always be replaced by the logical alternative of a number of irreducible
+equations. Reducible equations are therefore of subsidiary importance
+only; and we shall always suppose that the equation~\Eq{(1)} is irreducible.
+
+An algebraical function of~$x$ is regular except at a finite number
+of points which are \emph{poles} or \emph{branch points} of the function. Let $D$~be
+any closed simply connected domain in the plane of~$x$ which does
+not include any branch point. Then there are $n$~and only~$n$ distinct
+functions which are one-valued in~$D$ and satisfy the equation~\Eq{(1)}.
+These $n$~functions will be called the \emph{roots} of~\Eq{(1)} in~$D$. Thus if we
+write
+\[
+x = r (\cos \theta + i \sin \theta),
+\]
+where $-\pi < \theta \leq \pi$, then the roots of
+\[
+y^{2} - x = 0,
+\]
+in the domain
+\[
+0 < r_{1} \leq r \leq r_{2},\quad
+-\pi < -\pi + \delta \leq \theta \leq \pi - \delta < \pi,
+\]
+are $\sqrt{x}$~and~$-\sqrt{x}$, where
+\[
+\sqrt{x} = \sqrt{r} (\cos \tfrac{1}{2} \theta + i \sin \tfrac{1}{2} \theta).
+\]
+
+The relations which hold between the different roots of~\Eq{(1)} are of
+the greatest importance in the theory of functions\footnotemark.
+  \footnotetext{For fuller information the reader may be referred to Appell and Goursat's
+  \textit{Théorie des fonctions algébriques}.}%
+For our present
+purposes we require only the two which follow.
+
+\Item{(i)} Any symmetric polynomial in the roots~$y_{1}$, $y_{2}$,~\dots,~$y_{n}$ of~\Eq{(1)} is
+a rational function of~$x$.
+%% -----File: 016.png---
+
+\Item{(ii)} Any symmetric polynomial in~$y_{2}$, $y_{3}$, \dots, $y_{n}$ is a polynomial in~$y_{1}$
+with coefficients which are rational functions of~$x$.
+
+The first proposition follows directly from the equations
+\[
+\sum y_{1} y_{2} \dots y_{s} = (-1)^{s}(P_{n-s} / P_{0}) \quad   (s=1, 2, \dots, n).
+\]
+To prove the second we observe that
+\[
+\sum_{2, 3, \dots} y_{2} y_{3} \dots y_{s} = \sum_{1, 2, \dots} y_{1} y_{2} \dots y_{s-1} - y_{1}\sum_{2, 3, \dots} y_{2} y_{3} \dots y_{s-1}\DPtypo{.}{,}
+\]
+so that the theorem is true for $\sum y_{2} y_{3} \dots y_{s}$ if it is true for $\sum y_{2} y_{3} \dots y_{s-1}$.
+It is certainly true for
+\[
+y_{2} + y_{3} + \dots + y_{n} = (y_{1} + y_{2} + \dots + y_{n}) - y_{1}.
+\]
+It is therefore true for $\sum y_{2} y_{3} \dots y_{s}$, and so for any symmetric polynomial in
+$y_{2}, y_{3}, \dots, y_{n}$.
+
+\Paragraph{2}{4.} Elementary functions which are not rational or algebraical are
+called \emph{elementary transcendental functions} or elementary transcendents.
+They include all the remaining functions which are of ordinary occurrence
+in elementary analysis.
+
+The trigonometrical (or circular) and hyperbolic functions, direct
+and inverse, may all be expressed in terms of exponential or logarithmic
+functions by means of the ordinary formulae of elementary trigonometry.
+Thus, for example,
+\begin{gather*}
+\sin x = \frac{e^{ix} - e^{-ix}}{2i},\qquad  \sinh x = \frac{e^{x} - e^{-x}}{2},\\
+\arctan x =\frac{1}{2i} \log \left(\frac{1 + ix}{1 - ix}\right),\qquad  \argtanh x
+=\frac{1}{2}\log \left(\frac{1 + x}{1 - x}\right).
+\end{gather*}
+There was therefore no need to specify them particularly in our
+definition.
+
+The elementary transcendents have been further classified in a
+manner first indicated by Liouville\footnote{`Mémoire sur la classification des transcendantes, et sur l'impossibilité
+d'exprimer les racines de certaines équations en fonction finie explicite des
+coefficients', \textit{Journal de mathématiques}, ser.~1, vol.~2, 1837, pp.~56--104; `Suite du
+mémoire\dots', \ibid.\ vol.~3, 1838, pp. 523--546.}. According to him a function is
+a transcendent \emph{of the first order} if the signs of exponentiation or of
+the taking of logarithms which occur in the formula which defines
+it apply only to rational or algebraical functions. For example
+\[
+x e^{-x^{2}},\  e^{x^{2}} + e^{x}\sqrt{\log x}
+\]
+are of the first order; and so is
+\[
+\arctan \frac{y}{\sqrt{1 + x^{2}}} ,
+\]
+%% -----File: 017.png---
+where $y$ is defined by the equation
+\[
+y^{5} - y - x = 0;
+\]
+and so is the function $y$ defined by the equation
+\[
+y^{5} - y - e^{x} \log x = 0.
+\]
+
+An elementary transcendent \emph{of the second order} is one defined by
+a formula in which the exponentiations and takings of logarithms are
+applied to rational or algebraical functions or to transcendents of the
+first order. This class of functions includes many of great interest and
+importance, of which the simplest are
+\[
+e^{e^{x}},\: \log \log x.
+\]
+It also includes irrational and complex powers of $x$, since, \textit{e.g.},
+\[
+x^{\sqrt 2} = e^{\sqrt 2 \log x},\quad x^{1+i} = e^{(1+i) \log x};
+\]
+the function
+\[
+x^{x} = e^{x \log x};
+\]
+and the logarithms of the circular functions.
+
+It is of course presupposed in the definition of a transcendent of the
+second kind that the function in question is incapable of expression as
+one of the first kind or as a rational or algebraical function. The
+function
+\[
+e^{\log R(x)},
+\]
+where $R(x)$ is rational, is not a transcendent of the second kind, since
+it can be expressed in the simpler form $R(x)$.
+
+It is obvious that we can in this way proceed to define transcendents
+of the $n$th order for all values of $n$. Thus
+\[
+\log \log \log x,\; \log \log \log \log x, \dots
+\]
+are of the third, fourth, \dots\ orders.
+
+Of course a similar classification of algebraical functions can be and
+has been made. Thus we may say that
+\[
+\sqrt x,\; \sqrt {x + \sqrt x},\; \sqrt{ x + \sqrt {x + \sqrt x}}, \dots
+\]
+are algebraical functions of the first, second, third, \dots\ orders. But
+the fact that there is a general theory of algebraical equations and
+therefore of \emph{implicit} algebraical functions has deprived this classification
+of most of its importance. There is no such general theory
+of elementary transcendental equations\footnote
+  {The natural generalisations of the theory of algebraical equations are to be
+  found in parts of the theory of differential equations. See Königsberger,
+  `Bemerkungen zu Liouville's Classificirung der Transcendenten', \textit{Math.\ Annalen},
+  vol.~28, 1886, pp.~483--492.},
+and therefore we shall not
+%% -----File: 018.png---
+rank as `elementary' functions defined by transcendental equations
+such as
+\[
+y = x \log y,
+\]
+but incapable (as Liouville has shown that in this case $y$~is incapable)
+of explicit expression in finite terms.
+
+\Paragraph{2}{5.} The preceding analysis of elementary transcendental functions
+rests on the following theorems:
+
+\Item{(\ia)} $e^{x}$ is not an algebraical function of~$x$;
+
+\Item{(\ib)} $\log x$ is not an algebraical function of~$x$;
+
+\Item{(\ic)} $\log x$ is not expressible in finite terms by means of signs of
+exponentiation and of algebraical operations, explicit or implicit\footnotemark;
+  \footnotetext{For example, $\log x$~cannot be equal to~$e^{y}$, where $y$~is an algebraical function
+  of $x$.}
+
+\Item{(\id)} transcendental functions of the first, second, third,~\dots\ orders
+actually exist.
+
+A proof of the first two theorems will be given later, but limitations
+of space will prevent us from giving detailed proofs of the third and
+fourth. Liouville has given interesting extensions of some of these
+theorems: he has proved, for example, that no equation of the form
+\[
+Ae^{\alpha p} + Be^{\beta p} + \dots + Re^{\rho p} = S,
+\]
+where $p$, $A$, $B$,~\dots, $R$,~$S$ are algebraical functions of~$x$, and $\alpha$, $\beta$,~\dots, $\rho$
+different constants, can hold for all values of~$x$.
+
+\Section{III. The integration of elementary functions.
+Summary of results}
+
+In the following pages we shall be concerned exclusively with the
+problem of the integration of elementary functions. We shall endeavour
+to give as complete an account as the space at our disposal permits of
+the progress which has been made by mathematicians towards the
+solution of the two following problems:
+
+\Item{(i)} \begin{Result}if $f(x)$ is an elementary function, how can we determine
+whether its integral is also an elementary function?\end{Result}
+
+\Item{(ii)} \begin{Result}if the integral is an elementary function, how can we find it?\end{Result}
+
+It would be unreasonable to expect complete answers to these
+questions. But sufficient has been done to give us a tolerably complete
+insight into the nature of the answers, and to ensure that it
+%% -----File: 019.png---
+shall not be difficult to find the complete answers in any particular
+case which is at all likely to occur in elementary analysis or in its
+applications.
+
+It will probably be well for us at this point to summarise the
+principal results which have been obtained.
+
+\Paragraph{3}{1.} The integral of a rational function (\hyperlink{4 para 1.}{\textsc{iv.}})\ is \emph{always} an elementary
+function. It is either rational or the sum of a rational function and
+of a finite number of constant multiples of logarithms of rational
+functions (\hyperlink{4 para 1.}{\textsc{iv.},~1}).
+
+If certain constants which are the roots of an algebraical equation
+are treated as known then the form of the integral can always be
+determined completely. But as the roots of such equations are not in
+general capable of explicit expression in finite terms, it is not in
+general possible to express the integral in an absolutely explicit form
+(\hyperlink{4 para 2.}{\textsc{iv.};~2,~3}).
+
+We can always determine, by means of a finite number of
+the elementary operations of addition, multiplication, and division,
+whether the integral is rational or not. If it is rational, we can
+determine it completely by means of such operations; if not, we
+can determine its rational part (\hyperlink{4 para 4.}{\textsc{iv.};~4,~5}).
+
+The solution of the problem in the case of rational functions may
+therefore be said to be complete; for the difficulty with regard to the
+explicit solution of algebraical equations is one not of inadequate
+knowledge but of proved impossibility (\hyperlink{4 para 6.}{\textsc{iv.},~6}).
+
+\Paragraph{3}{2.} The integral of an algebraical function (\hyperlink{5 para 1.}{\textsc{v.}}), explicit or implicit,
+may or may not be elementary.
+
+If $y$ is an algebraical function of~$x$ then the integral $\int y\, dx$, or, more
+generally, the integral
+\[
+\int R(x,y)\, dx,
+\]
+where $R$~denotes a rational function, is, if an elementary function,
+either algebraical or the sum of an algebraical function and of a finite
+number of constant multiples of logarithms of algebraical functions.
+All algebraical functions which occur in the integral are \emph{rational
+functions of $x$~and~$y$} (\hyperlink{5 para 11.}{\textsc{v.}; 11--14},\hyperlink{5 para 18.}{~18}).
+
+These theorems give a precise statement of a general principle
+enunciated by Laplace\footnotemark:
+  \footnotetext{\textit{Théorie analytique des probabilités}, p.~7.}
+`\emph{l'intégrale d'une fonction différentielle
+%% -----File: 020.png---
+(algébrique) ne peut contenir d'autres quantités \DPtypo{radicaux}{radicales} que celles
+qui entrent dans cette fonction}'; and, we may add, cannot contain
+\emph{exponentials} at all. Thus it is impossible that
+\[
+\int \frac{dx}{\sqrt {1 + x^{2}}}
+\]
+should contain $e^{x}$ or~$\sqrt {1 - x}$: the appearance of these functions in
+the integral could only be apparent, and they could be eliminated
+before differentiation. Laplace's principle really rests on the fact, of
+which it is easy enough to convince oneself by a little reflection
+and the consideration of a few particular cases (though to give a
+rigorous proof is of course quite another matter), that \emph{differentiation
+will not eliminate exponentials or algebraical irrationalities}. Nor, we
+may add, will it eliminate logarithms except when they occur in the
+simple form
+\[
+A \log \phi (x),
+\]
+where $A$~is a constant, and this is why logarithms can only occur
+in this form in the integrals of rational or algebraical functions.
+
+We have thus a general knowledge of the form of the integral
+of an algebraical function~$y$, when it is itself an elementary
+function. Whether this is so or not of course depends on the nature
+of the equation $f(x, y)=0$ which defines~$y$. If this equation, when
+interpreted as that of a curve in the plane~$(x,y)$, represents a \emph{unicursal}
+curve, \textit{i.e.}~a curve which has the maximum number of double points
+possible for a curve of its degree, or whose \emph{deficiency} is zero, then
+$x$~and~$y$ can be expressed simultaneously as rational functions of a third
+variable~$t$, and the integral can be reduced by a substitution to that
+of a rational function (\hyperlink{5 para 2.}{\textsc{v.};~2}, \hyperlink{5 para 7.}{7--9}). In this case, therefore, the integral
+is always an elementary function. But this condition, though sufficient,
+is not necessary. It is in general true that, when $f(x, y)=0$ is not
+unicursal, the integral is not an elementary function but a new
+transcendent; and we are able to classify these transcendents according
+to the deficiency of the curve. If, for example, the deficiency is unity,
+then the integral is in general a transcendent of the kind known as
+\emph{elliptic integrals}, whose characteristic is that they can be transformed
+into integrals containing no other irrationality than the square root of
+a polynomial of the third or fourth degree (\hyperlink{5 para 20.}{\textsc{v.},~20}). But there are infinitely
+many cases in which the integral can be expressed by algebraical
+functions and logarithms. Similarly there are infinitely many cases
+in which integrals associated with curves whose deficiency is greater
+%% -----File: 021.png---
+than unity are in reality reducible to elliptic integrals. Such abnormal
+cases have formed the subject of many exceedingly interesting
+researches, but no general method has been devised by which we can
+always tell, after a finite series of operations, whether any given
+integral is really elementary, or elliptic, or belongs to a higher order
+of transcendents.
+
+When $f(x, y) = 0$ is unicursal we can carry out the integration
+completely in exactly the same sense as in the case of rational functions.
+In particular, if the integral is \emph{algebraical} then it can be found by
+means of elementary operations which are always practicable. And
+it has been shown, more generally, that we can always determine by
+means of such operations whether the integral of any given algebraical
+function is algebraical or not, and evaluate the integral when it is
+algebraical. And although the general problem of determining whether
+any given integral is an elementary function, and calculating it if it
+is one, has not been solved, the solution in the particular case in which
+the deficiency of the curve $f(x, y) = 0$ is unity is as complete as it is
+reasonable to expect any possible solution to be.
+
+\Paragraph{3}{3.} The theory of the integration of transcendental functions
+(\hyperlink{6 para 1.}{\textsc{vi.}})\ is naturally much less complete, and the number of classes
+of such functions for which general methods of integration exist is
+very small. These few classes are, however, of extreme importance
+in applications (\hyperlink{6 para 2.}{\textsc{vi.};~2,~3}).
+
+There is a general theorem concerning the form of an integral of
+a transcendental function, when it is itself an elementary function,
+which is quite analogous to those already stated for rational and
+algebraical functions. The general statement of this theorem will be
+found in \hyperlink{6 para 5.}{\textsc{vi.},~§5}; it shows, for instance, that the integral of a rational
+function of $x$,~$e^{x}$ and~$\log x$ is either a rational function of those
+functions or the sum of such a rational function and of a finite
+number of constant multiples of logarithms of similar functions.
+From this general theorem may be deduced a number of more precise
+results concerning integrals of more special forms, such as
+\[
+\int y e^{x}\, dx,\quad \int y \log x\, dx,
+\]
+where $y$~is an algebraical function of~$x$ (\hyperlink{6 para 4.}{\textsc{vi.};~4},\hyperlink{6 para 6.}{~6}).
+%% -----File: 022.png---
+
+\Section{IV. Rational functions}
+
+\Paragraph{4}{1.} It is proved in treatises on algebra\footnote
+  {See, \textit{e.g.}, Weber's \textit{Traité d'algèbre supérieure} (French translation by J.~Griess,
+  Paris, 1898), vol.~1, pp.~61--64, 143--149, 350--353; or Chrystal's \textit{Algebra}, vol.~1,
+  pp.~151--162.}
+that any polynomial
+\[
+Q(x) = b_{0}x^{n} + b_{1}x^{n-1} + \dots + b_{n}
+\]
+can be expressed in the form
+\[
+b_{0} (x - \alpha_{1})^{n_{1}}
+      (x - \alpha_{2})^{n_{2}} \dots
+      (x - \alpha_{r})^{n_{r}},
+\]
+where $n_{1}$, $n_{2}$,~\dots\ are positive integers whose sum is~$n$, and $\alpha_{1}$, $\alpha_{2}$,~\dots\ are
+constants; and that any rational function~$R(x)$, whose denominator
+is~$Q(x)$, may be expressed in the form
+\[
+A_{0}x^{p} + A_{1}x^{p-1} + \dots + A_{p}
+  + \sum_{s=1}^{r} \left\{
+    \frac{\beta_{s, 1}}{x - \alpha_{s}}
+  + \frac{\beta_{s, 2}}{(x - \alpha_{s})^{2}} + \dots
+  + \frac{\beta_{s, n_{s}}}{(x - \alpha_{s})^{n_{s}}}
+  \right\},
+\]
+where $A_{0}$, $A_{1}$,~\dots, $\beta_{s, 1}$,~\dots\ are also constants. It follows that
+\begin{multline*}
+\int R(x)\, dx
+  = A_{0} \frac{x^{p+1}}{p + 1}
+  + A_{1} \frac{x^{p}}{p} + \dots + A_{p} x + C \\
+  + \sum_{s=1}^{r} \left\{
+    \beta_{s, 1} \log(x - \alpha_{s})
+  - \frac{\beta_{s, 2}}{x - \alpha_{s}} - \dots
+  - \frac{\beta_{s, n_{s}}}{(n_{s} - 1)(x - \alpha_{s})^{n_{s} - 1}}
+  \right\}.
+\end{multline*}
+From this we conclude that \emph{the integral of any rational function is an
+elementary function which is rational save for the possible presence
+of logarithms of rational functions}. In particular the integral will be
+\emph{rational} if each of the numbers~$\beta_{s, 1}$ is zero: this condition is evidently
+necessary and sufficient. A necessary but not sufficient condition is
+that $Q(x)$~should contain no simple factors.
+
+The integral of the general rational function may be expressed in
+a very simple and elegant form by means of symbols of differentiation.
+We may suppose for simplicity that the degree of~$P(x)$ is less than
+that of~$Q(x)$; this can of course always be ensured by subtracting
+a polynomial from~$R(x)$. Then
+\begin{align*}
+R(x) &= \frac{P(x)}{Q(x)} \\
+     &= \frac{1}{(n_{1} - 1)! (n_{2} - 1)! \dots (n_{r} - 1)!}\:
+        \frac{\dd^{n - r}}{\dd \alpha_{1}^{n_{1} - 1}
+                           \dd \alpha_{2}^{n_{2} - 1} \dots
+                           \dd \alpha_{r}^{n_{r} - 1}}\:
+        \frac{P(x)}{Q_{0}(x)},
+\end{align*}
+where
+\[
+Q_{0}(x) = b_{0} (x - \alpha_{1}) (x - \alpha_{2}) \dots (x - \alpha_{r}).
+\]
+Now
+\[
+\frac{P(x)}{Q_{0}(x)}
+  = \varpi_{0}(x)
+  + \sum_{s=1}^{r} \frac{P(\alpha_{s})}{(x - \alpha_{s}) Q'_{0}(\alpha_{s})},
+\]
+%% -----File: 023.png---
+where $\varpi_{0}(x)$~is a polynomial; and so
+{\small \begin{multline*}
+\int R(x)\,dx =\\
+ \frac{1}{(n_{1}-1)! \dots(n_{r}-1)!}\:
+ \frac{\dd^{n-r}}{\dd \alpha_{1}^{n_{1}-1}\dots \dd \alpha_{r}^{n_{r}-1}}
+ \left[ \Pi_{0}(x) + \sum_{s=1}^{r} \frac{P(\alpha_{s})}{Q_{0}'(\alpha_{s})} \log (x - \alpha_{s}) \right],
+\end{multline*}}%
+where
+\[
+\Pi_{0}(x) = \int \varpi_{0}(x)\,dx.
+\]
+But
+\[
+\Pi(x) = \frac{\dd^{n - r} \Pi_{0}(x)}
+              {\dd \alpha_{1}^{n_{1} - 1}
+               \dd \alpha_{2}^{n_{2} - 1}\dots
+               \dd \alpha_{r}^{n_{r} - 1}}
+\]
+is also a polynomial, and the integral contains no polynomial term,
+since the degree of~$P(x)$ is less than that of~$Q(x)$. Thus $\Pi(x)$~must
+vanish identically, so that
+\begin{multline*}
+\int R(x)\,dx =\\
+  \frac{1}{(n_{1}-1)!\dots(n_{r}-1)!}\:
+  \frac{\dd^{n - r}}{\dd \alpha_{1}^{n_{1}-1}\dots\dd \alpha_{r}^{n_{r}-1}}
+  \left[\sum_{s=1}^{r} \frac{P(\alpha_{s})}{Q_{0}'(\alpha_{s})} \log (x - \alpha_{s}) \right].
+\end{multline*}
+
+For example
+\[
+\int\frac{dx}{\{(x-a)(x-b)\}^{2}}
+  = \frac{\dd^{2}}{\dd a\, \dd b}\left\{
+      \frac{1}{a-b} \log \left(\frac{x-a}{x-b}\right)
+    \right\}.
+\]
+
+That $\Pi_{0}(x)$~is annihilated by the partial differentiations performed on it
+may be verified directly as follows. We obtain~$\Pi_{0}(x)$ by picking out from
+the expansion
+\[
+\frac{P(x)}{x^{r}}
+  \left(1 + \frac{\alpha_{1}}{x} + \frac{\alpha_{1}^{2}}{x^{2}} + \dots\right)
+  \left(1 + \frac{\alpha_{2}}{x} + \frac{\alpha_{2}^{2}}{x^{2}} + \dots\right) \dots \dots
+\]
+the terms which involve positive powers of~$x$. Any such term is of the form
+\[
+Ax^{\nu-r-s_{1}-s_{2}-\dots} \alpha_{1}^{s_{1}} \alpha_{2}^{s_{2}} \dots,
+\]
+where
+\[
+s_{1} + s_{2} + \dots \leq \nu - r \leq m - r,
+\]
+$m$~being the degree of~$P$. It follows that
+\[
+s_{1} + s_{2} + \dots < n - r = (m_{1} - 1) + (m_{2} - 1) + \dots;
+\]
+so that at least one of $s_{1}$, $s_{2}$,~\dots\ must be less than the corresponding one of
+$m_{1} - 1$, $m_{2} - 1$,~\dots.
+
+It has been assumed above that if
+\[
+F(x, \alpha) = \int f(x, \alpha)\,dx,
+\]
+then
+\[
+\frac{\dd F}{\dd \alpha} = \int \frac{\dd f}{\dd \alpha}\,dx.
+\]
+%% -----File: 024.png---
+The first equation means that $f = \dfrac{\dd F}{\dd x}$ and the second that $\dfrac{\dd f}{\dd \alpha} = \dfrac{\dd^{2} F}{\dd x\,\dd \alpha}$. As it
+follows from the first that $\dfrac{\dd f}{\dd \alpha} = \dfrac{\dd^{2} F}{\dd \alpha\, \dd x}$, what has really been assumed is that
+\[
+\frac{\dd^{2} F}{\dd \alpha\, \dd x} = \frac{\dd^{2} F}{\dd x\, \dd \alpha}.
+\]
+It is known that this equation is always true for $x = x_{0}$, $\alpha = \alpha_{0}$ if a circle
+can be drawn in the plane of~$(x, \alpha)$ whose centre is~$(x_{0}, \alpha_{0})$ and within which
+the differential coefficients are continuous.
+
+\Paragraph{4}{2.} It appears from~§1 that the integral of a rational function is
+in general composed of two parts, one of which is a rational function
+and the other a function of the form
+\[
+\sum A \log(x - \alpha).
+\Tag{(1)}
+\]
+We may call these two functions the \emph{rational part} and the \emph{transcendental
+part} of the integral. It is evidently of great importance to
+show that the `transcendental part' of the integral is really transcendental
+and cannot be expressed, wholly or in part, as a rational or
+algebraical function.
+
+We are not yet in a position to prove this completely\footnotemark;
+  \footnotetext{The proof will be completed in \hyperlink{5 para 16.}{v.,~16.}}
+but we can
+take the first step in this direction by showing that \emph{no sum of the
+form~\Eq{(1)} can be rational, unless every~$A$ is zero}.
+
+Suppose, if possible, that
+\[
+\sum A \log(x - \alpha) = \frac{P(x)}{Q(x)},
+\Tag{(2)}
+\]
+where $P$~and~$Q$ are polynomials without common factor. Then
+\[
+\sum \frac{A}{x - \alpha} = \frac{P'Q - PQ'}{Q^{2}}.
+\Tag{(3)}
+\]
+
+Suppose now that $(x - p)^{r}$ is a factor of~$Q$. Then $P'Q - PQ'$ is
+divisible by~$(x-p)^{r-1}$ and by no higher power of~$x-p$. Thus the
+right-hand side of~\Eq{(3)}, when expressed in its lowest terms, has a factor
+$(x-p)^{r+1}$ in its denominator. On the other hand the left-hand side,
+when expressed as a rational fraction in its lowest terms, has no
+repeated factor in its denominator. Hence $r=0$, and so $Q$~is a constant.
+We may therefore replace~\Eq{(2)} by
+\[
+\sum A \log (x - \alpha) = P(x),
+\]
+and \Eq{(3)}~by
+\[
+\sum \frac{A}{x - \alpha} = P'(x).
+\]
+Multiplying by~$x-\alpha$, and making $x$~tend to~$\alpha$, we see that $A = 0$.
+%% -----File: 025.png---
+
+\Paragraph{4}{3.} The method of~§1 gives a complete solution of the problem if
+the roots of $Q(x)=0$ can be determined; and in practice this is
+usually the case. But this case, though it is the one which occurs
+most frequently in practice, is from a theoretical point of view an
+exceedingly special case. The roots of $Q(x)=0$ are not in general
+explicit algebraical functions of the coefficients, and cannot as a rule
+be determined in any explicit form. The method of partial fractions
+is therefore subject to serious limitations. For example, we cannot
+determine, by the method of decomposition into partial fractions, such
+an integral as
+\[
+\int \frac{4x^{9} + 21x^{6} + 2x^{3} - 3x^{2} - 3}
+          {(x^{7} - x + 1)^{2}}\, dx,
+\]
+or even determine whether the integral is rational or not, although it
+is in reality a very simple function. A high degree of importance
+therefore attaches to the further problem of determining the integral
+of a given rational function so far as possible in an absolutely explicit
+form and by means of operations which are always practicable.
+
+It is easy to see that a complete solution of this problem cannot be
+looked for.
+
+{\small Suppose for example that $P(x)$~reduces to unity, and that $Q(x)=0$ is
+an equation of the fifth degree, whose roots~$\alpha_{1}$, $\alpha_{2}$,~\dots\ $\alpha_{\DPtypo{6}{5}}$ are all distinct and
+not capable of explicit algebraical expression.
+
+Then
+\begin{align*}
+\int R(x) \,dx
+  &= \sum_{1}^{5} \frac{\log(x - \alpha_{s})}{Q'(\alpha_{s})}\\
+  &= \log \prod_{1}^{5} \left\{(x - \alpha_{s})^{1/Q'(\alpha_{s})}\right\},
+\end{align*}
+and it is only if at least two of the numbers~$Q'(\alpha_{s})$ are commensurable that
+any two or more of the factors $(x - \alpha_{s})^{1/Q'(\alpha_{s})}$ can be associated so as to give
+a single term of the type~$A \log S(x)$, where $S(x)$~is rational. In general this
+will not be the case, and so it will not be possible to express the integral in
+any finite form which does not explicitly involve the roots. A more precise
+result in this connection will be proved later~(§6).}
+
+\Paragraph{4}{4.} The first and most important part of the problem has been
+solved by Hermite, who has shown that the \emph{rational part} of the
+integral can always be determined without a knowledge of the roots of~$Q(x)$,
+and indeed without the performance of any operations other
+than those of elementary algebra\footnotemark.
+  \footnotetext{The following account of Hermite's method is taken in substance from
+  Goursat's \textit{Cours d'analyse mathématique} (first edition), t.~1, pp.~238--241.}
+%% -----File: 026.png---
+
+Hermite's method depends upon a fundamental theorem in
+elementary algebra\footnote{See Chrystal's \textit{Algebra}, vol.~1, pp.~119 \textit{et~seq.}}
+which is also of great importance in the ordinary
+theory of partial fractions, viz.:
+
+\begin{Result}
+`If $X_{1}$~and~$X_{2}$ are two polynomials in~$x$ which have no common
+factor, and $X_{3}$~any third polynomial, then we can determine two polynomials
+$A_{1}$,~$A_{2}$, such that
+\[
+A_{1}X_{1} + A_{2}X_{2} = X_{3}.\text{'}
+\]
+\end{Result}
+
+Suppose that
+\[
+Q(x) = Q_{1} Q_{2}^{2} Q_{3}^{3}\dots Q_{t}^{t},
+\]
+$Q_{1}$,~\dots\ denoting polynomials which have only simple roots and of
+which no two have any common factor. We can always determine
+$Q_{1}$,~\dots\ by elementary methods, as is shown in the elements of the
+theory of equations\footnotemark.
+  \footnotetext{See, for example, Hardy, \textit{A course of pure mathematics} (2nd~edition), p.~208.}
+
+We can determine $B$~and~$A_{1}$ so that
+\[
+B Q_{1} + A_{1} Q_{2}^{2} Q_{3}^{3}\dots Q_{t}^{t} = P,
+\]
+and therefore so that
+\[
+R(x)
+  = \frac{P}{Q}
+  = \frac{A_{1}}{Q_{1}} + \frac{B}{Q_{2}^{2} Q_{3}^{3}\dots Q_{t}^{t}}.
+\]
+By a repetition of this process we can express~$R(x)$ in the form
+\[
+\frac{A_{1}}{Q_{1}}
+  + \frac{A_{2}}{Q_{2}^{2}} + \dots
+  + \frac{A_{t}}{Q_{t}^{t}},
+\]
+and the problem of the integration of~$R(x)$ is reduced to that of the
+integration of a function
+\[
+\frac{A}{Q^{\nu}},
+\]
+where $Q$~is a polynomial whose roots are all distinct. Since this is so,
+$Q$~and its derived function~$Q'$ have no common factor: we can therefore
+determine $C$~and~$D$ so that
+\[
+CQ + DQ' = A.
+\]
+Hence
+\begin{align*}
+\int \frac{A}{Q^{\nu}}\,dx
+  &= \int \frac{CQ + DQ'}{Q^{\nu}}\,dx\\
+  &= \int \frac{C}{Q^{\nu-1}}\,dx
+   - \frac{1}{\nu - 1} \int D \frac{d}{dx} \left(\frac{1}{Q^{\nu-1}}\right) dx\\
+  &= - \frac{D}{(\nu - 1) Q^{\nu - 1}} + \int \frac{E}{Q^{\nu - 1}}\,dx,
+\end{align*}
+where
+\[
+E = C + \frac{D'}{\nu - 1}.
+\]
+%% -----File: 027.png---
+Proceeding in this way, and reducing by unity at each step the power
+of~$1/Q$ which figures under the sign of integration, we ultimately
+arrive at an equation
+\[
+\int \frac{A}{Q^{\nu}}\,dx = R_{\nu}(x) + \int \frac{S}{Q}\,dx,
+\]
+where $R_{\nu}$~is a rational function and $S$~a polynomial.
+
+The integral on the right-hand side has no rational part, since all
+the roots of~$Q$ are simple~(§2). Thus the rational part of $\int R(x)\,dx$ is
+\[
+R_{2}(x) + R_{3}(x) + \dots + R_{t}(x),
+\]
+and it has been determined without the need of any calculations other
+than those involved in the addition, multiplication and division of
+polynomials\footnotemark.
+  \footnotetext{The operation of forming the derived function of a given polynomial can of
+  course be effected by a combination of these operations.}
+
+\Paragraph{4}{5.} \Item{(i)} Let us consider, for example, the integral
+\[
+\int \frac{4x^{9} + 21x^{6} + 2x^{3} - 3x^{2} - 3}{(x^{7} - x + 1)^{2}}\,dx,
+\]
+mentioned above~(§3). We require polynomials $A_{1}$,~$A_{2}$ such that
+\[
+A_{1}X_{1} + A_{2}X_{2} = X_{3},
+\Tag{(1)}
+\]
+where
+\[
+X_{1} = x^{7} - x + 1,\quad
+X_{2} = 7x^{6} - 1,\quad
+X_{3} = 4x^{9} + 21x^{6} + 2x^{3} - 3x^{2} - 3.
+\]
+
+In general, if the degrees of $X_{1}$~and~$X_{2}$ are $m_{1}$~and~$m_{2}$, and that of~$X_{3}$
+does not exceed $m_{1} + m_{2} - 1$, we can suppose that the degrees of $A_{1}$~and~$A_{2}$ do
+not exceed $m_{2} - 1$ and $m_{1} - 1$ respectively. For we know that polynomials
+$B_{1}$~and~$B_{2}$ exist such that
+\[
+B_{1} X_{1} + B_{2} X_{2} = X_{3}.
+\]
+If $B_{1}$~is of degree not exceeding $m_{2} - 1$, we take $A_{1} = B_{1}$, and if it is of higher
+degree we write
+\[
+B_{1} = L_{1} X_{2} + A_{1},
+\]
+where $A_{1}$~is of degree not exceeding $m_{2} - 1$. Similarly we write
+\[
+B_{2} = L_{2} X_{1} + A_{2}.
+\]
+We have then
+\[
+(L_{1} + L_{2}) X_{1} X_{2} + A_{1} X_{1} + A_{2} X_{2} = X_{3}.
+\]
+In this identity $L_{1}$~or~$L_{2}$ or both may vanish identically, and in any case we
+see, by equating to zero the coefficients of the powers of~$x$ higher than the
+$(m_{1} + m_{2} - 1)$th, that $L_{1} + L_{2}$ vanishes identically. Thus $X_{3}$~is expressed in
+the form required.
+
+The actual determination of the coefficients in $A_{1}$~and~$A_{2}$ is most easily
+performed by equating coefficients. We have then $m_{1} + m_{2}$~linear equations
+%% -----File: 028.png---
+in the same number of unknowns. These equations must be consistent,
+since we know that a solution exists\footnotemark.
+  \footnotetext{It is easy to show that the solution is also unique.}
+
+If $X_{3}$~is of degree higher than $m_{1} + m_{2} - 1$, we must divide it by~$X_{1}X_{2}$ and
+express the remainder in the form required.
+
+In this case we may suppose $A_{1}$~of degree~$5$ and $A_{2}$~of degree~$6$, and we
+find that
+\[
+A_{1} = -3x^{2},\quad A_{2} = x^{3} + 3.
+\]
+Thus the rational part of the integral is
+\[
+- \frac{x^{3} + 3}{x^{7} - x + 1},
+\]
+and, since $-3x^{2} + (x^{3} + 3)'=0$, there is no transcendental part.
+
+\Item{(ii)} The following problem is instructive: \emph{to find the conditions that
+\[
+\int \frac{\alpha x^{2} + 2 \beta x + \gamma}{(Ax^{2} + 2Bx + C)^{2}}\, dx
+\]
+may be rational, and to determine the integral when it is rational.}
+
+We shall suppose that $Ax^{2} + 2Bx + C$ is not a perfect square, as if it were
+the integral would certainly be rational. We can determine $p$,~$q$ and~$r$
+so that
+\[
+p(Ax^{2} + 2Bx + C) + 2(qx + r)(Ax + B) = \alpha x^{2} + 2 \beta x + \gamma,
+\]
+and the integral becomes
+\begin{multline*}
+p \int\frac{dx}{Ax^{2} + 2Bx + C}
+  - \int(qx + r)\, \frac{d}{dx} \left(\frac{1}{Ax^{2} + 2Bx + C}\right) dx\\
+  = -\frac{qx + r}{Ax^{2} + 2Bx + C}
+    + (p + q) \int\frac{dx}{Ax^{2} + 2Bx + C}.
+\end{multline*}
+The condition that the integral should be rational is therefore $p + q = 0$.
+
+Equating coefficients we find
+\[
+A(p + 2q) = \alpha, \quad
+B(p + q) + Ar = \beta, \quad
+Cp + 2Br = \gamma.
+\]
+Hence we deduce
+\[
+p = -\frac{\alpha}{A}, \quad
+q =  \frac{\alpha}{A}, \quad
+r =  \frac{\beta}{A},
+\]
+and $A \gamma + C \alpha = 2B \beta$. The condition required is therefore that the two quadratics
+$\alpha x^{2} + 2 \beta x + \gamma$ and $Ax^{2} + 2Bx + C$ should be harmonically related, and in this
+case
+\[
+\int \frac{\alpha x^{2} + 2 \beta x + \gamma}{(Ax^{2} + 2Bx + C)^{2}}\,dx
+  = -\frac{\alpha x + \beta}{A (Ax^{2} + 2Bx + C)}.
+\]
+
+\Item{(iii)} Another method of solution of this problem is as follows. If we write
+\[
+Ax^{2} + 2Bx + C = A(x - \lambda)(x - \mu),
+\]
+and use the bilinear substitution
+\[
+x = \frac{\lambda y + \mu}{y + 1},
+\]
+then the integral is reduced to one of the form
+\[
+\int \frac{ay^{2} + 2by + c}{y^{2}}\,dy,
+\]
+%% -----File: 029.png---
+and is rational if and only if $b = 0$. But this is the condition that the
+quadratic $ay^{2} + 2by + c$, corresponding to $\alpha x^{2} + 2 \beta x + \gamma$, should be harmonically
+related to the degenerate quadratic~$y$, corresponding to $Ax^{2} + 2Bx + C$. The
+result now follows from the fact that harmonic relations are not changed by
+bilinear transformation.
+
+It is not difficult to show, by an adaptation of this method, that
+\[
+\int\frac{(\alpha x^{2} + 2 \beta x + \gamma)
+          (\alpha_{1} x^{2} + 2 \beta_{1} x + \gamma_{1}) \dots
+          (\alpha_{n} x^{2} + 2 \beta_{n} x + \gamma_{n})}
+         {(Ax^{2} + 2Bx + C)^{n+2}}\, dx
+\]
+is rational if all the quadratics are harmonically related to any one of those
+in the numerator. This condition is sufficient but not necessary.
+
+\Item{(iv)} As a further example of the use of the method~(ii) the reader may
+show that \begin{Result}the necessary and sufficient condition that
+\[
+\int\frac{f(x)}{\{ F(x) \}^{2}}\,dx,
+\]
+where $f$~and~$F$ are polynomials with no common factor, and $F$~has no repeated
+factor, should be rational, is that $f'F' - fF''$ should be divisible by~$F$.\end{Result}
+
+\Paragraph{4}{6.} It appears from the preceding paragraphs that we can always
+find the rational part of the integral, and can find the complete integral
+if we can find the roots of $Q(x) = 0$. The question is naturally
+suggested as to the maximum of information which can be obtained
+about the logarithmic part of the integral in the general case in which
+the factors of the denominator cannot be determined explicitly. For
+there are polynomials which, although they cannot be completely resolved
+into such factors, can nevertheless be partially resolved. For example
+\begin{multline*}
+x^{14} - 2x^{8} - 2x^{7} - x^{4} - 2x^{3} + 2x + 1
+  = (x^{7} + x^{2} - 1) (x^{7} - x^{2} - 2x - 1),\\
+\shoveleft{x^{14} - 2x^{8} - 2x^{7} - 2x^{4} - 4x^{3} - x^{2} + 2x + 1}\\
+  = \{x^{7} + x^{2} \sqrt{2} + x (\sqrt{2} - 1) - 1\}
+    \{x^{7} - x^{2} \sqrt{2} - x (\sqrt{2} + 1) - 1\}.
+\end{multline*}
+The factors of the first polynomial have rational coefficients: in the
+language of the theory of equations, the polynomial is \emph{reducible in the
+rational domain}. The second polynomial is reducible in the domain
+formed by the \emph{adjunction} of the single irrational~$\sqrt{2}$ to the rational
+domain\footnotemark.
+  \footnotetext{See Cajori, \textit{An introduction to the modern theory of equations} (Macmillan,
+  1904); Mathews, \textit{Algebraic equations} (\textit{Cambridge tracts in mathematics}, no.~6),
+  pp.~6--7.}
+
+We may suppose that every possible decomposition of~$Q(x)$ of this
+nature has been made, so that
+\[
+Q = Q_{1} Q_{2}\dots Q_{t}.
+\]
+%% -----File: 030.png---
+Then we can resolve~$R(x)$ into a sum of partial fractions of the type
+\[
+\int\frac{P_{\nu}}{Q_{\nu}}\, dx,
+\]
+and so we need only consider integrals of the type
+\[
+\int\frac{P}{Q}\, dx,
+\]
+where no further resolution of~$Q$ is possible or, in technical language,
+\emph{$Q$~is irreducible by the adjunction of any algebraical irrationality}.
+
+Suppose that this integral can be evaluated in a form involving only
+constants which can be expressed explicitly in terms of the constants
+which occur in~$P/Q$. It must be of the form
+\[
+A_{1} \log X_{1} + \dots + A_{k} \log X_{k},
+\Tag{(1)}
+\]
+where the~$A$'s are constants and the~$X$'s polynomials. We can
+suppose that no~$X$ has any repeated factor~$\xi^{m}$, where $\xi$~is a polynomial.
+For such a factor could be determined rationally in terms of the coefficients
+of~$X$, and the expression~\Eq{(1)} could then be modified by
+taking out the factor~$\xi^{m}$ from~$X$ and inserting a new term $mA \log \xi$.
+And for similar reasons we can suppose that no two~$X$'s have any
+factor in common.
+
+Now
+\[
+\frac{P}{Q}
+  = A_{1}\frac{X'_{1}}{X_{1}}
+  + A_{2}\frac{X'_{2}}{X_{2}} + \dots
+  + A_{k}\frac{X'_{k}}{X_{k}},
+\]
+or
+\[
+P X_{1} X_{2} \dots X_{k}
+  = Q \sum A_{\nu} X_{1} \dots X_{\nu-1} X'_{\nu} X_{\nu+1} \dots X_{k}.
+\]
+All the terms under the sign of summation are divisible by~$X_{1}$ save the
+first, which is prime to~$X_{1}$. Hence $Q$~must be divisible by~$X_{1}$: and
+similarly, of course, by $X_{2}$, $X_{3}$,~\dots, $X_{k}$. But, since $P$~is prime to~$Q$,
+$X_{1} X_{2} \dots X_{k}$ is divisible by~$Q$. Thus $Q$~must be a constant multiple of
+$X_{1} X_{2} \dots X_{k}$. But $Q$~is \textit{ex~hypothesi} not resoluble into factors which
+contain only explicit algebraical irrationalities. Hence all the~$X$'s
+save one must reduce to constants, and so $P$~must be a constant
+multiple of~$Q'$, and
+\[
+\int\frac{P}{Q}\, dx = A \log Q,
+\]
+where $A$~is a constant. Unless this is the case the integral cannot be
+expressed in a form involving only constants expressed explicitly in
+terms of the constants which occur in $P$~and~$Q$.
+
+{\small Thus, for instance, the integral
+\[
+\int\frac{dx}{x^{5} + ax + b}
+\]
+%% -----File: 031.png---
+cannot, except in special cases\footnotemark,
+  \footnotetext{The equation $x^{5} + ax + b = 0$ is soluble by radicals in certain cases. See
+  Mathews, \textit{l.c.}, pp.~52~\textit{et~seq.}}
+be expressed in a form involving only
+constants expressed explicitly in terms of $a$~and~$b$; and the integral
+\[
+\int \frac{5x^{4} + c}{x^{5} + ax + b}\, dx
+\]
+can in general be so expressed if and only if $c = a$. We thus confirm an
+inference made before~(§3) in a less accurate way.
+
+Before quitting this part of our subject we may consider one further
+problem: \emph{under what circumstances is
+\[
+\int R(x)\, dx = A \log R_{1}(x)
+\]
+where $A$~is a constant and $R_{1}$~rational?} Since the integral has no rational
+part, it is clear that $Q(x)$~must have only simple factors, and that the degree
+of~$P(x)$ must be less than that of~$Q(x)$. We may therefore use the formula
+\[
+\int R(x)\,dx
+  = \log \prod_{1}^{r} \left\{
+      (x - \alpha_{s})^{P(\alpha_{s})/Q'(\alpha_{s})}
+  \right\}.
+\]
+The necessary and sufficient condition is that all the numbers $P(\alpha_{s})/Q'(\alpha_{s})$
+should be commensurable. If~\textit{e.g.}
+\[
+R(x) = \frac{x - \gamma}{(x - \alpha) (x - \beta)},
+\]
+{\Loosen then $(\alpha - \gamma)/(\alpha - \beta)$ and $(\beta - \gamma)/(\beta - \alpha)$ must be commensurable, \textit{i.e.}\ $(\alpha - \gamma)/(\beta - \gamma)$
+must be a rational number. If the denominator is given we can find all the
+values of~$\gamma$ which are admissible: for $\gamma = (\alpha q - \beta p)/(q - p)$, where $p$~and~$q$ are
+integers.}
+
+\Paragraph{4}{7.} Our discussion of the integration of rational functions is now
+complete. It has been throughout of a theoretical character. We
+have not attempted to consider what are the simplest and quickest
+methods for the actual calculation of the types of integral which occur
+most commonly in practice. This problem lies outside our present
+range: the reader may consult
+
+O. Stolz, \textit{Grundzüge der \DPtypo{Differential- und Integralrechnung}{Differential- und~Integralrechnung}}, vol.~1,
+ch.~7:
+
+J. Tannery, \textit{Leçons d'algèbre et d'analyse,} vol.~2, ch.~18:
+
+Ch.-J. de~la Vallée-Poussin, \textit{Cours d'analyse,} ed.~3, vol.~1, ch.~5:
+
+T.~J.~I'A. Bromwich, \textit{Elementary integrals} (Bowes and Bowes,
+1911):
+
+G.~H. Hardy, \textit{A course of pure mathematics,} ed.~2, ch.~6.
+%% -----File: 032.png---
+
+\Section{V. Algebraical Functions}
+
+\Paragraph{5}{1.} We shall now consider the integrals of algebraical functions,
+explicit or implicit. The theory of the integration of such functions is
+far more extensive and difficult than that of rational functions, and
+we can give here only a brief account of a few of the most important
+results and of the most obvious of their applications.
+
+If $y_{1}$, $y_{2}$,~\dots, $y_{n}$ are algebraical functions of~$x$, then any algebraical
+function~$z$ of $x$,~$y_{1}$, $y_{2}$,~\dots, $y_{n}$ is an algebraical function of~$x$. This is
+obvious if we confine ourselves to \emph{explicit} algebraical functions. In
+the general case we have a number of equations of the type
+\[
+P_{\nu, 0}(x) y_{\nu}^{m_{\nu}}
+  + P_{\nu, 1}(x) y_{\nu}^{m_{\nu}-1} + \dots
+  + P_{\nu, m_{\nu}}(x) = 0 \quad(\nu = 1, 2,\dots, n),
+\]
+and
+\[
+P_{0}(x, y_{1},\dots, y_{n}) z^{m} + \dots
+  + P_{m}(x, y_{1},\dots, y_{n}) = 0,
+\]
+where the~$P$'s represent polynomials in their arguments. The elimination
+of $y_{1}$, $y_{2}$,~\dots, $y_{n}$ between these equations gives an equation in~$z$
+whose coefficients are polynomials in $x$~only.
+
+The importance of this from our present point of view lies in the
+fact that we may consider the standard algebraical integral under any
+of the forms
+\[
+\int y\,dx,
+\]
+where $f(x, y) = 0$;
+\[
+\int R(x, y)\,dx,
+\]
+where $f(x, y) = 0$ and $R$~is rational; or
+\[
+\int R(x, y_{1},\dots, y_{n})\,dx,
+\]
+where $f_{1}(x, \DPtypo{y}{y_1}) = 0$,~\dots, $f_{n}(x, y_{n}) = 0$. It is, for example, much more
+convenient to treat such an irrational as
+\[
+\frac{x - \sqrt{x+1} - \sqrt{x-1}}
+     {1 + \sqrt{x+1} + \sqrt{x-1}}
+\]
+as a rational function of $x$,~$y_{1}$,~$y_{2}$, where $y_{1} = \sqrt{x+1}$, $y_{2} = \sqrt{x-1}$,
+$y_{1}^{2} = x + 1$, $y_{2}^{2} = x- 1$, than as a rational function of $x$~and~$y$, where
+\begin{gather*}
+y = \sqrt{x+1} + \sqrt{x-1}, \\
+y^{4} - 4xy^{2} + 4 = 0.
+\end{gather*}
+To treat it as a simple irrational~$y$, so that our fundamental equation is
+\[
+(x - y)^{4} - 4x(x - y)^{2} (1 + y)^{2} + 4(1 + y)^{4} = 0
+\]
+is evidently the least convenient course of all.
+%% -----File: 033.png---
+
+Before we proceed to consider the general form of the integral of an
+algebraical function we shall consider one most important case in which
+the integral can be at once reduced to that of a rational function, and
+is therefore always an elementary function itself.
+
+\Paragraph{5}{2.} The class of integrals alluded to immediately above is that
+covered by the following theorem.
+
+\begin{Result}
+If there is a variable~$t$ connected with $x$ and $y$ (or $y_{1}$, $y_{2}$, \dots, $y_{n}$)
+by rational relations
+\[
+x = R_{1}(t),\quad y = R_{2}(t)
+\]
+(or $y_{1} = R_{2}^{(1)}(t)$, $y_{2} = R_{2}^{(2)}(t)$, \dots), then the integral
+\[
+\int R(x, y)\, dx
+\]
+(or $\int R(x, y_{1}, \dots, y_{n})\, dx$) is an elementary function.
+\end{Result}
+
+The truth of this proposition follows immediately from the
+equations
+\begin{gather*}
+R(x, y) = R\{R_{1}(t), R_{2}(t)\} = S(t), \\
+\frac{dx}{dt} = R_{1}'(t) = T(t), \\
+\int R(x, y)\, dx = \int S(t) T(t)\, dt = \int U(t)\, dt,
+\end{gather*}
+where all the capital letters denote rational functions.
+
+The most important case of this theorem is that in which $x$ and $y$
+are connected by the general quadratic relation
+\[
+(a, b, c, f, g, h \between x, y, 1)^{2} = 0.
+\]
+The integral can then be made rational in an infinite number of ways\DPtypo{}{.}
+For suppose that $(\xi, \eta)$ is any point on the conic, and that
+\[
+(y - \eta) = t(x - \xi)
+\]
+is any line through the point. If we eliminate $y$ between these
+equations, we obtain an equation of the second degree in $x$, say
+\[
+T_{0}x^{2} + 2T_{1}x + T_{2} = 0,
+\]
+where $T_{0}$, $T_{1}$, $T_{2}$ are polynomials in $t$. But one root of this equation
+must be $\xi$, which is independent of $t$; and when we divide by $x - \xi$ we
+obtain an equation of the \emph{first} degree for the abscissa of the variable
+point of intersection, in which the coefficients are again polynomials
+in $t$. Hence this abscissa is a rational function of $t$; the ordinate of
+the point is also a rational function of $t$, and as $t$ varies this point
+%% -----File: 034.png---
+coincides with every point of the conic in turn. In fact the equation
+of the conic may be written in the form
+\[
+au^{2} + 2huv + bv^{2} + 2(a\xi + h\eta + g)u + 2(h\xi + b\eta + f)v = 0,
+\]
+where $u = x - \xi$, $v = y - \eta$, and the other point of intersection of the line
+$v = tu$ and the conic is given by
+\begin{gather*}
+x = \xi - \frac{2\{a\xi + h\eta + g + t(h\xi + b\eta + f)\}}{a + 2ht +bt^{2}}, \\
+y = \eta - \frac{2t\{a\xi + h\eta + g + t(h\xi + b\eta + f)\}}{a + 2ht +bt^{2}}.
+\end{gather*}
+
+An alternative method is to write
+\[
+ax^{2} + 2hxy + by^{2} = b(y - \mu x)(y - \mu' x),
+\]
+so that $y - \mu x = 0$ and $y - \mu' x = 0$ are parallel to the asymptotes of
+the conic, and to put
+\[
+y - \mu x = t.
+\]
+Then
+\[
+y - \mu' x = - \frac{2gx + 2fy + c}{bt};
+\]
+and from these two equations we can calculate $x$ and $y$ as rational
+functions of $t$. The principle of this method is of course the same as
+that of the former method: $(\xi, \eta)$ is now at infinity, and the pencil of
+lines through $(\xi, \eta)$ is replaced by a pencil parallel to an asymptote.
+
+The most important case is that in which $b = -1$, $f = h = 0$, so that
+\[
+y^{2} = ax^{2} + 2gx + c.
+\]
+The integral is then made rational by the substitution
+\[
+x = \xi - \frac{2(a\xi + g - t\eta)}{a - t^{2}},\quad y = \eta - \frac{2t(a\xi + g - t\eta)}{a - t^{2}}
+\]
+where $\xi$, $\eta$ are any numbers such that
+\[
+\eta^{2} = a\xi^{2} + 2g\xi + c.
+\]
+We may for instance suppose that $\xi = 0$, $\eta = \sqrt c$; or that $\eta = 0$, while $\xi$
+is a root of the equation $a\xi^{2} + 2g\xi + c = 0$. Or again the integral is
+made rational by putting $y - x \sqrt a = t$, when
+\[
+x = - \frac{t^{2} - c}{2(t \sqrt a - g)},\quad y = \frac{(t^{2} + c) \sqrt a - 2gt}{2(t \sqrt a - g)}.
+\]
+
+\Paragraph{5}{3.} We shall now consider in more detail the problem of the calculation of
+\[
+\int R(x, y)\, dx,
+\]
+where
+\[
+y = \sqrt X = \sqrt{ax^{2} + 2bx + c}\;\footnote
+  {We now write $b$ for $g$ for the sake of symmetry in notation.}
+\]
+%% -----File: 035.png---
+The most interesting case is that in which $a$,~$b$,~$c$ and the constants which
+occur in~$R$ are real, and we shall confine our attention to this case.
+
+Let
+\[
+R(x, y) = \frac{P(x, y)}{Q(x, y)},
+\]
+where $P$~and~$Q$ are polynomials. Then, by means of the equation
+\[
+y^{2} = ax^{2} + 2bx + c,
+\]
+$R(x, y)$ may be reduced to the form
+\[
+\frac{A + B \sqrt{X}}{C + D \sqrt{X}}
+  = \frac{(A + B \sqrt{X})(C - D \sqrt{X})}{C^{2} - D^{2}X},
+\]
+where $A$, $B$, $C$,~$D$ are polynomials in~$x$; and so to the form $M + N \sqrt{X}$, where
+$M$~and~$N$ are rational, or (what is the same thing) the form
+\[
+P + \frac{Q}{\sqrt{X}},
+\]
+where $P$~and~$Q$ are rational. The rational part may be integrated by the
+methods of \hyperlink{6 para 1.}{section~\textsc{iv.}}, and the integral
+\[
+\int \frac{Q}{\sqrt{X}}\, dx
+\]
+may be reduced to the sum of a number of integrals of the forms
+\[
+\int \frac{x^{r}}{\sqrt{X}}\, dx,\quad
+\int \frac{dx}{(x - p)^{r} \sqrt{X}},\quad
+\int \frac{\xi x + \eta}{(\alpha x^{2} + 2\beta x + \gamma)^{r} \sqrt{X}}\, dx,
+\Tag{(1)}
+\]
+where $p$, $\xi$,~$\eta$, $\alpha$,~$\beta$,~$\gamma$ are real constants and $r$~a positive integer. The result
+is generally required in an explicitly real form: and, as further progress
+depends on transformations involving~$p$ (or $\alpha$,~$\beta$,~$\gamma$), it is generally not
+advisable to break up a quadratic factor $\alpha x^{2} + 2\beta x + \gamma$ into its constituent
+linear factors when these factors are complex.
+
+All of the integrals~\Eq{(1)} may be reduced, by means of elementary formulae
+of reduction\footnotemark,
+  \footnotetext{See, for example, Bromwich, \textit{l.c.}, pp.~16~\textit{et~seq.}}
+to dependence upon three fundamental integrals, viz.
+\[
+\int \frac{dx}{\sqrt{X}},\quad
+\int \frac{dx}{(x - p) \sqrt{X}},\quad
+\int \frac{\xi x + \eta}{(\alpha x^{2} + 2\beta x + \gamma) \sqrt{X}}\, dx.
+\Tag{(2)}
+\]
+
+\Paragraph{5}{4.} The first of these integrals may be reduced, by a substitution of the
+type $x = t + k$, to one or other of the three standard forms
+\[
+\int \frac{dt}{\sqrt{m^{2} - t^{2}}},\quad
+\int \frac{dt}{\sqrt{t^{2} + m^{2}}},\quad
+\int \frac{dt}{\sqrt{t^{2} - m^{2}}},
+\]
+where $m > 0$. These integrals may be rationalised by the substitutions
+\[
+t = \frac{2mu}{1 + u^{2}},\quad
+t = \frac{2mu}{1 - u^{2}},\quad
+t = \frac{m(1 + u^{2})}{2u};
+\]
+but it is simpler to use the transcendental substitutions
+\[
+t = m \sin \phi,\quad
+t = m \sinh \phi,\quad
+t = m \cosh \phi.
+\]
+%% -----File: 036.png---
+These last substitutions are generally the most convenient for the reduction
+of an integral which contains one or other of the irrationalities
+\[
+\sqrt{m^{2} - t^{2}},\quad
+\sqrt{t^{2} + m^{2}},\quad
+\sqrt{t^{2} - m^{2}},
+\]
+though the alternative substitutions
+\[
+t = m \tanh \phi,\quad
+t = m \tan \phi,\quad
+t = m \sec \phi
+\]
+are often useful.
+
+It has been pointed out by Dr~Bromwich that the forms usually given in
+text-books for these three standard integrals, viz.
+\[
+\arcsin \frac{t}{m},\quad
+\argsinh \frac{t}{m},\quad
+\argcosh \frac{t}{m}
+\]
+are not quite accurate. It is obvious, for example, that the first two of these
+functions are odd functions of~$m$, while the corresponding integrals are even
+functions. The correct formulae are
+\[
+\arcsin \frac{t}{|m|},\quad
+\argsinh \frac{t}{|m|} = \log \frac{t + \sqrt{t^{2} + m^{2}}}{|m|}
+\]
+and
+\[
+±\argcosh \frac{|t|}{|m|}
+  = \log \left|\frac{t + \sqrt{t^{2} - m^{2}}}{m}\right|,
+\]
+where the ambiguous sign is the same as that of~$t$. It is in some ways more
+convenient to use the equivalent forms
+\[
+\arctan \frac{t}{\sqrt{m^{2} - t^{2}}},\quad
+\argtanh \frac{t}{\sqrt{t^{2} + m^{2}}},\quad
+\argtanh \frac{t}{\sqrt{t^{2} - m^{2}}}.
+\]
+
+\Paragraph{5}{5.} The integral
+\[
+\int \frac{dx}{(x - p) \sqrt{X}}
+\]
+may be evaluated in a variety of ways.
+
+If $p$~is a root of the equation $X = 0$, then $X$~may be written in the form
+$a(x - p)(x - q)$, and the value of the integral is given by one or other of the
+formulae
+\begin{gather*}
+\int \frac{dx}{(x - p) \sqrt{(x - p)(x - q)}}
+  = \frac{2}{q - p} \sqrt{\frac{x - q}{x - p}}, \\
+\int \frac{dx}{(x - p)^{5/2}} = -\frac{2}{3(x - p)^{3/2}}.
+\end{gather*}
+We may therefore suppose that $p$~is not a root of $X = 0$.
+
+\Item{(i)} We may follow the general method described above, taking
+\[
+\xi = p,\quad
+\eta = \sqrt{ap^{2} + 2bp + c}\footnotemark.
+\]
+\footnotetext{Cf.\ Jordan, \textit{Cours d'analyse}, ed.~2, vol.~2, p.~21.}%
+Eliminating~$y$ from the equations
+\[
+y^{2} = ax^{2} + 2bx + c,\quad
+y - \eta = t(x - \xi),
+\]
+and dividing by $x - \xi$, we obtain
+\[
+t^{2}(x - \xi) + 2\eta t - a(x + \xi) - 2b = 0,
+\]
+and so
+\[
+-\frac{2\, dt}{t^{2} - a}
+  = \frac{dx}{t(x - \xi) + \eta}
+  = \frac{dx}{y}.
+\]
+%% -----File: 037.png---
+
+Hence
+\[
+\int \frac{dx}{(x - \xi)y} = -2 \int \frac{dt}{(x - \xi)(t^{2} - a)}.
+\]
+But
+\[
+(t^{2} - a)(x - \xi) = 2a\xi + 2b - 2\eta t;
+\]
+and so
+\begin{align*}
+\int \frac{dx}{(x - p)y}
+   &= - \int \frac{dt}{a\xi + b - \eta t}
+   = \frac{1}{\eta} \log(a\xi + b - \eta t)\\
+  &= \frac{1}{\sqrt{ap^{2} + 2bp + c}} \log\{t \sqrt{ap^{2} + 2bp + c} - ap -b\}.
+\end{align*}
+If $ap^{2} + 2bp + c < 0$ the transformation is imaginary.
+
+Suppose, \textit{e.g.}, (\ia)~$y = \sqrt{x + 1}$, $p = 0$, or (\ib)~$y = \sqrt{x - 1}$, $p = 0$. We find
+
+\Item{(\ia)}
+\[
+\int \frac{dx}{x \sqrt{x + 1}} = \log(t - \tfrac{1}{2}),
+\]
+where
+\[
+t^{2}x + 2t - 1 = 0,
+\]
+or
+\[
+t = \frac{-1 + \sqrt{x + 1}}{x};
+\]
+and
+
+\Item{(\ib)}
+\[
+\int \frac{dx}{x \sqrt{x - 1}} = -i \log(it - \tfrac{1}{2}),
+\]
+where
+\[
+t^{2}x + 2it - 1 = 0.
+\]
+Neither of these results is expressed in the simplest form, the second in
+particular being very inconvenient.
+
+\Item{(ii)} The most straightforward method of procedure is to use the
+substitution
+\[
+x - p = \frac{1}{t}.
+\]
+We then obtain
+\[
+\int \frac{dx}{(x - p)y}
+  = \int \frac{dt}{\sqrt{a_{1}t^{2} + 2b_{1}t + c_{1}}},
+\]
+where $a_{1}$,~$b_{1}$,~$c_{1}$ are certain simple functions of $a$,~$b$,~$c$, and~$p$. The further
+reduction of this integral has been discussed already.
+
+\Item{(iii)} A third method of integration is that adopted by Sir~G. Greenhill\footnotemark,
+  \footnotetext{A.~G. Greenhill, \textit{A chapter in the integral calculus} (Francis Hodgson, 1888),
+  p.~12: \textit{Differential and integral calculus}, p.~399.}%
+who uses the transformation
+\[
+t = \frac{\sqrt{ax^{2} + 2bx + c}}{x - p}\DPtypo{}{.}
+\]
+It will be found that
+\[
+\int \frac{dx}{(x - p) \sqrt{X}}
+  = \int \frac{dt}{\sqrt{(ap^{2} + 2bp + c)t^{2} + b^{2} - ac}},
+\]
+which is of one of the three standard forms mentioned in \hyperlink{5 para 4.}{§~4}.
+%% -----File: 038.png---
+
+\Paragraph{5}{6.} It remains to consider the integral
+\[
+\int \frac{\xi x + \eta}{(\alpha x^{2} + 2\beta x + \gamma) \sqrt{X}}\, dx
+  = \int \frac{\xi x + \eta}{X_{1} \sqrt{X}}\, dx,
+\]
+where $\alpha x^{2} + 2\beta x + \gamma$ or~$X_{1}$ is a quadratic with complex linear factors. Here
+again there is a choice of methods at our disposal.
+
+We may suppose that $X_{1}$~is not a constant multiple of~$X$. If it is, then
+the value of the integral is given by the formula
+\[
+\int \frac{\xi x + \eta}{(ax^{2} + 2bx + c)^{3/2}}\, dx
+  = \frac{\eta(ax + b) - \xi(bx + c)}{\sqrt{(ac - b^{2})(ax^{2} + 2bx + c)}}\footnotemark.
+\]
+\footnotetext{Bromwich, \textit{l.c.}, p.~16.}
+
+\Item{(i)} The standard method is to use the substitution
+\[
+x = \frac{\mu t + \nu}{t + 1},
+\Tag{(1)}
+\]
+where $\mu$~and~$\nu$ are so chosen that
+\[
+a\mu\nu + b(\mu + \nu) + c = 0,\quad
+\alpha\mu\nu + \beta(\mu + \nu) + \gamma = 0.
+\Tag{(2)}
+\]
+
+The values of $\mu$~and~$\nu$ which satisfy these conditions are the roots of the
+quadratic
+\[
+(a\beta - b\alpha)\mu^{2} - (c\alpha - a\gamma)\mu + (b\gamma - c\beta) = 0.
+\]
+The roots will be real and distinct if
+\[
+(c\alpha - a\gamma)^{2} > 4(a\beta - b\alpha)(b\gamma - c\beta),
+\]
+or if
+\[
+(a\gamma + c\alpha - 2b\beta)^{2} > 4(ac - b^{2})(\alpha\gamma - \beta^{2}).
+\Tag{(3)}
+\]
+Now $\alpha\gamma - \beta^{2} > 0$, so that \Eq{(3)}~is certainly satisfied if $ac - b^{2} < 0$.   But if $ac - b^{2}$
+and $\alpha\gamma - \beta^{2}$ are both positive then $a\gamma$~and~$c\alpha$ have the same sign, and
+\begin{align*}
+(a\gamma + c\alpha - 2b\beta)^{2}
+  &\geq (|a\gamma + c\alpha| - 2|b\beta|)^{2} > 4\{\sqrt{ac\alpha\gamma} - |b\beta|\}^{2}\\
+  &= 4[(ac - b^{2})(\alpha\gamma - \beta^{2}) + \{|b| \sqrt{\alpha\gamma} - |\beta| \sqrt{ac}\}^{2}]\\
+  &\geq 4(ac - b^{2})(\alpha\gamma - \beta^{2}).
+\end{align*}
+Thus the values of $\mu$~and~$\nu$ are in any case real and distinct.
+
+It will be found, on carrying out the substitution~\Eq{(1)}, that
+\[
+\int \frac{\xi x + \eta}{X_{1} \sqrt{X}}\, dx
+  = H \int \frac{t\,dt}{(\mathbf{A}t^{2} + \mathbf{B}) \sqrt{At^{2} + B}}
+  + K \int \frac{dt}{(\mathbf{A}t^{2} + \mathbf{B}) \sqrt{At^{2} + B}},
+\]
+where $\mathbf{A}$,~$\mathbf{B}$, $A$,~$B$, $H$, and~$K$ are constants. Of these two integrals, the first
+is rationalised by the substitution
+\[
+\frac{1}{\sqrt{At^{2} + B}} = u,
+\]
+and the second by the substitution
+\[
+\frac{t}{\sqrt{At^{2} + B}} = v.\footnote
+  {The method sketched here is that followed by Stolz (see the references given
+on p.~21). Dr~Bromwich's method is different in detail but the same in principle.}
+\]
+
+It should be observed that this method fails in the special case in which
+%% -----File: 039.png---
+$a\beta - b\alpha = 0$. In this case, however, the substitution $ax + b = t$ reduces the
+integral to one of the form
+\[
+\int \frac{Ht + K}{(\mathbf{A}t^{2} + \mathbf{B}) \sqrt{At^{2} + B}} \,dt,
+\]
+and the reduction may then be completed as before.
+
+\Item{(ii)} An alternative method is to use Sir~G. Greenhill's substitution
+\[
+t = \sqrt{\frac{X}{\alpha x^{2} + 2\beta x + \gamma}}
+  = \sqrt{\frac{X}{X_{1}}}.
+\]
+If
+\[
+J = (a\beta - b\alpha)x^{2} - (c\alpha - a\gamma)x + (b\gamma - c\beta),
+\]
+then
+\[
+\frac{1}{t}\, \frac{dt}{dx} = \frac{J}{XX_{1}}.
+\Tag{(1)}
+\]
+The maximum and minimum values of~$t$ are given by $J = 0$.
+
+Again
+\[
+t^{2} - \lambda
+  = \frac{(a - \lambda\alpha)x^{2} + 2(b - \lambda\beta)x + (c - \lambda\gamma)}{X_{1}};
+\]
+and the numerator will be a perfect square if
+\[
+K = (\alpha\gamma - \beta^{2})\lambda^{2}
+  - (a\gamma + c\alpha - 2b\beta)\lambda
+  + (ac - b^{2}) = 0.
+\]
+
+It will be found by a little calculation that the discriminant of this
+quadratic and that of~$J$ differ from one another and from
+\[
+(\phi - \phi_{1})(\phi - \phi_{1}')(\phi' - \phi_{1})(\phi' - \phi_{1}'),
+\]
+where $\phi$,~$\phi'$ are the roots of $X = 0$ and $\phi_{1}$,~$\phi_{1}'$ those of $X_{1} = 0$, only by
+a constant factor which is always negative. Since $\phi_{1}$~and~$\phi_{1}'$ are conjugate
+complex numbers, this product is positive, and so $J = 0$ and $K = 0$ have real
+roots\footnotemark.
+  \footnotetext{That the roots of $J = 0$ are real has been proved already (\hyperlink{5 para 6.}{p.~28}) in a different
+  manner.}%
+We denote the roots of the latter by
+\[
+\lambda_{1}, \lambda_{2} \quad(\lambda_{1} > \lambda_{2}).
+\]
+Then
+\begin{alignat*}{2}
+\lambda_{1} - t^{2}
+  &=  \frac{\{x \sqrt{\lambda_{1}\alpha - a} + \sqrt{\lambda_{1}\gamma - c}\}^{2}}{X_{1}}
+  &&= \frac{(mx + n)^{2}}{X_{1}},
+\Tag{(2)}\\
+%
+t^{2} - \lambda_{2}
+  &=  \frac{\{x \sqrt{a - \lambda_{2}\alpha} + \sqrt{c - \lambda_{2}\gamma}\}^{2}}{X_{1}}
+  &&= \frac{(m'x + n')^{2}}{X_{1}},
+\Tag{(2')}
+\end{alignat*}
+say. Further, since $t^{2} - \lambda$ can vanish for two equal values of~$x$ only if $\lambda$~is
+equal to $\lambda_{1}$~or~$\lambda_{2}$, \textit{i.e.}\ when $t$~is a maximum or a minimum, $J$~can differ from
+\[
+(mx + n)(m'x + n')
+\]
+only by a constant factor; and by comparing coefficients and using the
+identity
+\[
+(\lambda_{1}\alpha - a)(a - \lambda_{2}\alpha)
+  = \frac{(a\beta - b\alpha)^{2}}{\alpha\gamma - \beta^{2}},
+\]
+we find that
+\[
+J = \sqrt{\alpha\gamma - \beta^{2}}(mx + n)(m'x + n').
+\Tag{(3)}
+\]
+
+Finally, we can write $\xi x + \eta$ in the form
+\[
+A(mx + n) + B(m'x + n').
+\]
+%% -----File: 040.png---
+{\small
+Using equations \Eq{(1)}, \Eq{(2)}, \Eq{(2')}, and \Eq{(3)}, we find that
+\begin{align*}
+\int \frac{\xi x + \eta}{X_{1} \sqrt{X}}\, dx &= \int \frac{A(mx + n) + B(m'x + n')}{J} \sqrt{X_{1}}\, dt\\
+  &= \frac{A}{\sqrt{\alpha\gamma - \beta^{2}}} \int \frac{dt}{\sqrt{\lambda_{1} - t^{2}}} + \frac{B}{\sqrt{\alpha\gamma - \beta^{2}}} \int \frac{dt}{\sqrt{t^{2} - \lambda_{2}}},
+\end{align*}
+and the integral is reduced to a sum of two standard forms.
+
+This method is very elegant, and has the advantage that the whole work
+of transformation is performed in one step. On the other hand it is
+somewhat artificial, and it is open to the logical objection that it introduces
+the root $\sqrt{X_{1}}$, which, in virtue of Laplace's principle (\hyperlink{3 para 2.}{\textsc{iii}., 2}), cannot really
+be involved in the final result\footnotemark.\footnotetext
+{The superfluous root may be eliminated from the result by a trivial transformation,
+just as $\sqrt{1 + x^{2}}$ may be eliminated from
+\[
+\arcsin \frac{x}{\sqrt{1 + x^{2}}}
+\]
+by writing this function in the form $\arctan x$.}}
+
+\Paragraph{5}{7.} We may now proceed to consider the general case to which the
+theorem of \hyperlink{4 para 2.}{\textsc{iv}., 2} applies. It will be convenient to recall two well-known
+definitions in the theory of algebraical plane curves. A curve
+of degree $n$ can have at most $\frac{1}{2} (n - 1)(n - 2)$ double points\footnotemark.\footnotetext
+{Salmon, \emph{Higher plane curves}, p.~29.}
+If the
+actual number of double points is $\nu$, then the number
+\[
+p = \tfrac{1}{2} (n - 1)(n - 2) - \nu
+\]
+is called the \emph{deficiency}\footnote{Salmon, \ibid., p.~29. French \emph{genre}, German \emph{Geschlecht}.} of the curve.
+
+If the coordinates $x$, $y$ of the points on a curve can be expressed
+\emph{rationally} in terms of a parameter $t$ by means of equations
+\[
+x = R_{1}(t),\quad y = R_{2}(t),
+\]
+then we shall say that the curve is \emph{unicursal}. In this case we have
+seen that we can always evaluate
+\[
+\int R(x, y)\, dx
+\]
+in terms of elementary functions.
+
+The fundamental theorem in this part of our subject is
+
+\begin{Result}
+`A curve whose deficiency is zero is unicursal, and vice versa'.
+\end{Result}
+
+{\Loosen Suppose first that the curve possesses the maximum number of
+double points\footnotemark.\footnotetext
+  {We suppose in what follows that the singularities of the curve are all ordinary
+nodes. The necessary modifications when this is not the case are not difficult to
+make. An ordinary multiple point of order~$k$ may be regarded as equivalent to
+$\frac{1}{2} k(k - 1)$ ordinary double points. A curve of degree~$n$ which has an ordinary
+multiple point of order~$n - 1$, equivalent to $\frac{1}{2} (n - 1)(n - 2)$ ordinary double points,
+is therefore unicursal. The theory of higher plane curves abounds in puzzling
+particular cases which have to be fitted into the general theory by more or less
+obvious conventions, and to give a satisfactory account of a complicated compound
+singularity is sometimes by no means easy. In the investigation which follows we
+confine ourselves to the simplest case.}
+Since
+\[
+\tfrac{1}{2} (n - 1)(n - 2) + n - 3 = \tfrac{1}{2} (n - 2)(n + 1) - 1,
+\]
+%% -----File: 041.png---
+and $\frac{1}{2} (n - 2)(n + 1)$ points are just sufficient to determine a curve of
+degree $n - 2$\footnotemark,
+  \footnotetext{Salmon, \textit{l.c.}, p.~16.}
+we can draw, through the $\frac{1}{2} (n - 1)(n - 2)$ double points
+and $n - 3$~other points chosen arbitrarily on the curve, a simply infinite
+set of curves of degree~$n - 2$, which we may suppose to have the
+equation
+\[
+g(x, y) + t\,h(x, y) = 0,
+\]
+where $t$~is a variable parameter and $g = 0$, $h = 0$ are the equations of
+two particular members of the set. Any one of these curves meets
+the given curve in $n(n - 2)$~points, of which $(n - 1)(n - 2)$ are accounted
+for by the $\frac{1}{2} (n - 1)(n - 2)$ double points, and $n - 3$~by the
+other $n - 3$ arbitrarily chosen points. These
+\[
+(n - 1)(n - 2) + n - 3 = n(n - 2) - 1
+\]
+points are independent of~$t$; and so there is but \emph{one} point of intersection
+which depends on~$t$. The coordinates of this point are given by
+\[
+g(x, y) + t\,h(x, y) = 0,\quad f(x, y) = 0.
+\]
+The elimination of~$y$ gives an equation of degree $n(n - 2)$ in~$x$, whose
+coefficients are polynomials in~$t$; and but one root of this equation
+varies with~$t$. The eliminant is therefore divisible by a factor of
+degree $n(n - 2) - 1$ which does not contain~$t$. There remains a simple
+equation in~$x$ whose coefficients are polynomials in~$t$. Thus the
+$x$-coordinate of the variable point is determined as a rational function
+of~$t$, and the $y$-coordinate may be similarly determined.
+
+We may therefore write
+\[
+x = R_{1}(t),\quad y = R_{2}(t).
+\]
+If we reduce these fractions to the same denominator, we express the
+coordinates in the form
+\[
+x = \frac{\phi_{1}(t)}{\phi_{3}(t)},\quad
+y = \frac{\phi_{2}(t)}{\phi_{3}(t)},
+\Tag{(1)}
+\]
+where $\phi_{1}$,~$\phi_{2}$,~$\phi_{3}$ are polynomials which have no common factor. The
+polynomials will in general be of degree~$n$; none of them can be of
+%% -----File: 042.png---
+higher degree, and one at least must be actually of that degree, since
+an arbitrary straight line
+\[
+\lambda x + \mu y + \nu = 0
+\]
+must cut the curve in exactly $n$~points\footnotemark.
+  \footnotetext{See Niewenglowski's \textit{Cours de géométrie analytique}, vol.~2, p.~103. By way of
+  illustration of the remark concerning particular cases in the footnote~(§) to \hyperlink{5 para 7.}{page~30},
+  the reader may consider the example given by Niewenglowski in which
+  \[
+  x = \frac{t^{2}}{t^{2} - 1},\quad
+  y = \frac{t^{2} + 1}{t^{2} - 1};
+  \]
+  equations which appear to represent the straight line $2x = y + 1$ (part of the line
+  only, if we consider only real values of~$t$).}
+
+We can now prove the second part of the theorem. If
+\[
+x : y : 1 :: \phi_{1}(t) : \phi_{2}(t) : \phi_{3}(t),
+\]
+where $\phi_{1}$,~$\phi_{2}$,~$\phi_{3}$ are polynomials of degree~$n$, then the line
+\[
+ux + vy + w = 0
+\]
+will meet the curve in $n$~points whose parameters are given by
+\[
+u\phi_{1}(t) + v\phi_{2}(t) + w\phi_{3}(t) = 0.
+\]
+
+This equation will have a double root~$t_{0}$ if
+\begin{align*}
+  u\phi_{1}(t_{0}) + v\phi_{2}(t_{0}) + w\phi_{3}(t_{0}) &= 0,\\
+  u\phi_{1}'(t_{0}) + v\phi_{2}'(t_{0}) + w\phi_{3}'(t_{0}) &= 0.
+\end{align*}
+Hence the equation of the tangent at the point~$t_{0}$ is
+\[
+\begin{vmatrix}
+  x  &  y  &  1\\
+  \phi_{1}(t_{0}) & \phi_{2}(t_{0}) & \phi_{3}(t_{0})\\
+  \phi_{1}'(t_{0})& \phi_{2}'(t_{0})& \phi_{3}'(t_{0})
+\end{vmatrix}
+= 0.
+\Tag{(2)}
+\]
+
+If $(x, y)$~is a fixed point, then the equation~\Eq{(2)} may be regarded as
+an equation to determine the parameters of the points of contact
+of the tangents from~$(x, y)$. Now
+\[
+\phi_{2}(t_{0})\phi_{3}'(t_{0}) - \phi_{2}'(t_{0})\phi_{3}(t_{0})
+\]
+is of degree $2n - 2$ in~$t_{0}$, the coefficient of~$t_{0}^{2n-1}$ obviously vanishing.
+Hence in general the number of tangents which can be drawn to a
+unicursal curve from a fixed point (the \emph{class} of the curve) is~$2n - 2$.
+But the class of a curve whose only singular points are $\delta$~nodes is
+known\footnote
+  {Salmon, \textit{l.c.}, p.~54.}
+to be $n(n - 1) - 2\delta$. Hence the number of nodes is
+\[
+\tfrac{1}{2} \{n(n - 1) - (2n - 2)\} = \tfrac{1}{2} (n - 1)(n - 2).
+\]
+
+It is perhaps worth pointing out how the proof which precedes requires
+modification if some only of the singular points are nodes and the rest
+ordinary cusps. The first part of the proof remains unaltered. The equation~\Eq(2)
+%% -----File: 043.png---
+must now be regarded as giving the values of~$t$ which correspond to
+(\ia)~points at which the tangent passes through~$(x, y)$ and (\ib)~cusps, since any
+line through a cusp `cuts the curve in two coincident points'\footnotemark.
+  \footnotetext{This means of course that the equation obtained by substituting for $x$~and~$y$,
+  in the equation of the line, their parametric expressions in terms of~$t$, has a
+  repeated root. This property is possessed by the tangent at an ordinary point and
+  by any line through a cusp, but not by any line through a node except the two
+  tangents.}
+We have
+therefore
+\[
+2n - 2 = m + \kappa,
+\]
+where $m$~is the class of the curve. But
+\[
+m = n(n - 1) - 2\delta - 3\kappa,\footnotemark
+\]
+\footnotetext{Salmon, \textit{l.c.}, p.~65.}%
+and so
+\[
+\delta + \kappa = \tfrac{1}{2}(n - 1)(n - 2).\footnotemark
+\]
+\footnotetext{I owe this remark to Mr~A.~B. Mayne. Dr~Bromwich has however pointed
+  out to me that substantially the same argument is given by Mr~W.~A. Houston, `Note
+  on unicursal plane curves', \textit{Messenger of mathematics}, vol.~28, 1899, pp.~187--189.}%
+
+
+\Paragraph{5}{8.} \Item{(i)} The preceding argument fails if $n < 3$, but we have already
+seen that all conics are unicursal. The case next in importance is
+that of a cubic with a double point. If the double point is not at
+infinity we can, by a change of origin, reduce the equation of the
+curve to the form
+\[
+(ax + by)(cx + dy) = px^{3} + 3qx^{2}y + 3rxy^{2} + sy^{3};
+\]
+and, by considering the intersections of the curve with the line
+$y = tx$, we find
+\[
+x = \frac{(a + bt)(c + dt)}{p + 3qt + 3rt^{2} + st^{3}},\quad
+y = \frac{t(a + bt)(c + dt)}{p + 3qt + 3rt^{2} + st^{3}}.
+\]
+
+If the double point is at infinity, the equation of the curve is of the
+form
+\[
+(\alpha x + \beta y)^{2}(\gamma x + \delta y) + \epsilon x + \zeta y + \theta = 0,
+\]
+the curve having a pair of parallel asymptotes; and, by considering
+the intersection of the curve with the line $\alpha x + \beta y = t$, we find
+\[
+x = -\frac{\delta t^{3} + \zeta t + \beta\theta}
+          {(\beta\gamma - \alpha\delta)t^{2} + \epsilon\beta - \alpha\zeta},\quad
+y =  \frac{\gamma t^{3} + \epsilon t + \alpha\theta}
+          {(\beta\gamma - \alpha\delta)\DPtypo{t^{3}}{t^{2}} + \epsilon\beta - \alpha\zeta}.
+\]
+
+\Item{(ii)} The case next in complexity is that of a quartic with three double
+points.
+
+\Item{(\ia)} The lemniscate
+\[
+(x^{2} + y^{2})^{2} = a^{2}(x^{2} - y^{2})
+\]
+has three double points, the origin and the circular points at infinity. The
+circle
+\[
+x^{2} + y^{2} = t(x - y)
+\]
+%% -----File: 044.png---
+passes through these points and one other fixed point at the origin, as it
+touches the curve there. Solving, we find
+\[
+x = \frac{a^{2}t(t^{2} + a^{2})}{t^{4} + a^{4}},\quad
+y = \frac{a^{2}t(t^{2} - a^{2})}{t^{4} + a^{4}}.
+\]
+
+\Item{(\ib)} The curve
+\[
+2ay^{3} - 3a^{2}y^{2} = x^{4} - 2a^{2}x^{2}
+\]
+has the double points $(0, 0)$, $(a, a)$, $(-a, a)$. Using the auxiliary conic
+\[
+x^{2} - ay = tx(y - a),
+\]
+we find
+\[
+x = \frac{a}{t^{3}}(2 - 3t^{2}),\quad
+y = \frac{a}{2t^{4}}(2 - 3t^{2})(2 - t^{2}).
+\]
+
+\Item{(iii)} \Item{(\ia)} The curve
+\[
+y^{n} = x^{n} + ax^{n-1}
+\]
+has a multiple point of order~$n - 1$ at the origin, and is therefore unicursal.
+In this case it is sufficient to consider the intersection of the curve with the
+line $y = tx$. This may be harmonised with the general theory by regarding
+the curve
+\[
+y^{n-3}(y - tx) = 0,
+\]
+as passing through each of the $\frac{1}{2} (n - 1)(n - 2)$ double points collected at the
+origin and through $n - 3$~other fixed points collected at the point
+\[
+x = -a,\quad y = 0.
+\]
+
+The curves
+\begin{align*}
+y^{n} &= x^{n} + ax^{n-1},
+\Tag{(1)} \\
+y^{n} &= 1 + az,
+\Tag{(2)}
+\end{align*}
+are projectively equivalent, as appears on rendering their equations homogeneous
+by the introduction of variables $z$ in~\Eq{(1)} and $x$ in~\Eq{(2)}. We conclude
+that \Eq{(2)}~is unicursal, having the maximum number of double points at
+infinity. In fact we may put
+\[
+y = t,\quad az = t^{n} - 1.
+\]
+The integral
+\[
+\int R\{z, \sqrt[n]{1 + az}\}\,dz
+\]
+is accordingly an elementary function.
+
+\Item{(\ib)} The curve
+\[
+y^{m} = A(x - a)^{\mu} (x - b)^{\nu}
+\]
+is unicursal if and only if either (i)~$\mu = 0$ or (ii)~$\nu = 0$ or (iii)~$\mu + \nu = m$.
+Hence the integral
+\[
+\int R\{x, (x - a)^{\mu/m} (x - b)^{\nu/n}\}\,dx
+\]
+is an elementary function, for all forms of~$R$, in these three cases only; of
+course it is integrable for special forms of~$R$ in other cases\footnotemark.
+  \footnotetext{See Ptaszycki, `Extrait d'une lettre adressée à M.~Hermite', \textit{Bulletin des
+  sciences mathématiques}, ser.~2, vol.~12, 1888, pp.~262--270: Appell and Goursat,
+  \textit{Théorie des fonctions algébriques}, p.~245.}
+%% -----File: 045.png---
+
+\Paragraph{5}{9.} There is a similar theory connected with unicursal curves
+in space of any number of dimensions. Consider for example the
+integral
+\[
+\int R\{x, \sqrt{ax + b}, \sqrt{cx + d}\}\,dx.
+\]
+A linear substitution $\DPtypo{x}{y} = lx + m$ reduces this integral to the form
+\[
+\int R_{1}\{y, \sqrt{y + 2}, \sqrt{y - 2}\}\,dy;
+\]
+and this integral can be rationalised by putting
+\[
+y = t^{2} + \frac{1}{t^{2}},\quad
+\sqrt{y + 2} = t + \frac{1}{t},\quad
+\sqrt{y - 2} = t - \frac{1}{t}.
+\]
+
+The curve whose Cartesian coordinates $\xi$,~$\eta$,~$\zeta$ are given by
+\[
+\xi : \eta : \zeta : 1 :: t^{4} + 1 : t(t^{2} + 1) : t(t^{2} - 1) : t^{2},
+\]
+is a unicursal twisted quartic, the intersection of the parabolic cylinders
+\[
+\xi = \eta^{2} - 2,\quad
+\xi = \zeta^{2} + 2.
+\]
+
+It is easy to deduce that the integral
+\[
+\int R\left\{x, \sqrt{\frac{ax + b}{mx + n}}, \sqrt{\frac{cx + d}{mx + n}}\right\} dx
+\]
+is always an elementary function.
+
+\Paragraph{5}{10.} When the deficiency of the curve $f(x, y) = 0$ is not zero, the
+integral
+\[
+\int R(x, y)\,dx
+\]
+is in general not an elementary function; and the consideration of
+such integrals has consequently introduced a whole series of classes of
+new transcendents into analysis. The simplest case is that in which
+the deficiency is unity: in this case, as we shall see later on, the
+integrals are expressible in terms of elementary functions and certain
+new transcendents known as elliptic integrals. When the deficiency
+rises above unity the integration necessitates the introduction of new
+transcendents of growing complexity.
+
+But there are infinitely many particular cases in which integrals,
+associated with curves whose deficiency is unity or greater than unity,
+%% -----File: 046.png---
+can be expressed in terms of elementary functions, or are even
+algebraical themselves. For instance the deficiency of
+\[
+y^{2} = 1 + x^{3}
+\]
+is unity. But
+\begin{gather*}
+\int\frac{x + 1}{x - 2}\, \frac{dx}{\sqrt{1 + x^{3}}}
+  = 3\log\frac{(1 + x)^{2} - 3\sqrt{1 + x^{3}}}
+              {(1 + x)^{2} + 3\sqrt{1 + x^{3}}},\\
+%
+\int\frac{2 - x^{3}}{1 + x^{3}}\, \frac{dx}{\sqrt{1 + x^{3}}}
+  = \frac{2x}{\sqrt{1 + x^{3}}}.
+\end{gather*}
+And, before we say anything concerning the new transcendents to
+which integrals of this class in general give rise, we shall consider what
+has been done in the way of formulating rules to enable us to identify
+such cases and to assign the form of the integral when it is an
+elementary function. It will be as well to say at once that this
+problem has not been solved completely.
+
+\Paragraph{5}{11.} The first general theorem of this character deals with the
+case in which the integral is algebraical, and asserts that \begin{Result}if
+\[
+u = \int y\, dx
+\]
+is an algebraical function of~$x$, then it is a rational function of~$x$~and~$y$.
+\end{Result}
+
+Our proof will be based on the following lemmas.
+
+\begin{Result}
+\Item{(1)} If $f(x, y)$ and~$g(x, y)$ are polynomials, and there is no factor
+common to all the coefficients of the various powers of~$y$ in~$g(x, y)$; and
+\[
+f(x, y) = g(x, y)h(x),
+\]
+where $h(x)$~is a rational function of~$x$; then $h(x)$~is a polynomial.
+\end{Result}
+
+Let $h = P/Q$, where $P$~and~$Q$ are polynomials without a common
+factor. Then
+\[
+fQ = gP.
+\]
+If $x - a$~is a factor of~$Q$, then
+\[
+g(a, y) = 0
+\]
+for all values of~$y$; and so all the coefficients of powers of~$y$ in~$g(x, y)$
+are divisible by~$x - a$, which is contrary to our hypotheses. Hence
+$Q$~is a constant and $h$~a polynomial.
+
+\begin{Result}
+\Item{(2)} Suppose that $f(x, y)$~is an irreducible polynomial, and that
+$y_{1}$, $y_{2}$,~\dots, $y_{n}$ are the roots of
+\[
+f(x, y) = 0
+\]
+%% -----File: 047.png---
+in a certain domain~$D$. Suppose further that $\phi(x, y)$~is another
+polynomial, and that
+\[
+\phi(x, y_{1}) = 0.
+\]
+Then
+\[
+\phi(x, y_{s}) = 0,
+\]
+where $y_{s}$~is any one of the roots of~\Eq{(1)}; and
+\[
+\phi(x, y) = f(x, y) \psi(x, y),
+\]
+where $\psi(x, y)$ also is a polynomial in $x$~and~$y$.
+\end{Result}
+
+Let us determine the highest common factor~$\varpi$ of $f$~and~$\phi$, considered
+as polynomials in~$y$, by the ordinary process for the determination
+of the highest common factor of two polynomials. This
+process depends only on a series of algebraical divisions, and so $\varpi$~is a
+polynomial in~$y$ with coefficients rational in~$x$. We have therefore
+\begin{gather*}
+\varpi(x, y) = \omega(x, y) \lambda(x),
+\Tag{(1)}\\
+f(x, y) = \omega(x, y) p(x, y) \mu(x) = g(x, y) \mu(x),
+\Tag{(2)}\\
+\phi(x, y) = \omega(x, y) q(x, y) \nu(x) = h(x, y) \nu(x),
+\Tag{(3)}
+\end{gather*}
+where $\omega$, $p$,~$q$,~$g$, and~$h$ are polynomials and $\lambda$,~$\mu$, and~$\nu$ rational
+functions; and evidently we may suppose that neither in~$g$ nor in~$h$
+have the coefficients of all powers of~$y$ a common factor. Hence, by
+Lemma~(1), $\mu$~and~$\nu$ are polynomials. But $f$~is irreducible, and therefore
+$\mu$ and either $\omega$~or~$p$ must be constants. If $\omega$~were a constant,
+$\varpi$~would be a function of $x$~only. But this is impossible. For we can
+determine polynomials $L$,~$M$ in~$y$, with coefficients rational in~$x$, such
+that
+\[
+Lf + M\phi = \varpi,
+\Tag{(4)}
+\]
+and the left-hand side of~\Eq{(4)} vanishes when we write $y_{1}$ for~$y$. Hence
+$p$~is a constant, and so $\omega$~is a constant multiple of~$f$. The truth of
+the lemma now follows from~\Eq{(3)}.
+
+It follows from Lemma~(2) that \begin{Result}$y$~cannot satisfy any equation of
+degree less than~$n$ whose coefficients are polynomials in~$x$.
+\end{Result}
+
+\begin{Result}
+\Item{(3)} If $y$~is an algebraical function of~$x$, defined by an equation
+\[
+f(x, y) = 0
+\Tag{(1)}
+\]
+of degree~$n$, then any rational function $R(x, y)$ of $x$~and~$y$ can be
+expressed in the form
+\[
+R(x, y) = R_{0} + R_{1}y + \dots + R_{n-1}y^{n-1},
+\Tag{(2)}
+\]
+where $R_{0}$, $R_{1}$,~\dots, $R_{n-1}$ are rational functions of~$x$.
+\end{Result}
+%% -----File: 048.png---
+
+The function~$y$ is one of the $n$~roots of~\Eq{(1)}. Let $y$, $y'$, $y''$,~\dots\ be the
+complete system of roots. Then
+\begin{align*}
+R(x, y) &= \frac{P(x, y)}{Q(x, y)}\\
+        &= \frac{P(x, y) Q(x, y') Q(x, y'')\dots}
+                {Q(x, y) Q(x, y') Q(x, y'')\dots},
+\Tag{(3)}
+\end{align*}
+where $P$~and~$Q$ are polynomials. The denominator is a polynomial in~$x$
+whose coefficients are symmetric polynomials in $y$, $y'$, $y''$,~\dots, and is
+therefore, by \hyperlink{2 para 3.}{\textsc{ii.},~§3,~(i)}, a rational function of~$x$. On the other hand
+\[
+Q(x, y') Q(x, y'')\dots
+\]
+is a polynomial in~$x$ whose coefficients are symmetric polynomials
+in $y'$, $y''$,~\dots, and therefore, by \hyperlink{2 para 3.}{\textsc{ii.},~§3,~(ii)}, polynomials in~$y$ with
+coefficients rational in~$x$. Thus the numerator of~\Eq{(3)} is a polynomial
+in~$y$ with coefficients rational in~$x$.
+
+It follows that $R (x, y)$~is a polynomial in~$y$ with coefficients rational
+in~$x$. From this polynomial we can eliminate, by means of~\Eq{(1)}, all
+powers of~$y$ as high as or higher than the~$n$th. Hence $R(x, y)$~is of
+the form prescribed by the lemma.
+
+\Paragraph{5}{12.} We proceed now to the proof of our main theorem. We have
+\[
+\int y\,dx = u
+\]
+where $u$~is algebraical. Let
+\[
+f(x, y) = 0,\quad \psi(x, u) = 0
+\Tag{(1)}
+\]
+be the irreducible equations satisfied by $y$~and~$u$, and let us suppose
+that they are of degrees $n$~and~$m$ respectively. The first stage in the
+proof consists in showing that
+\[
+m = n.
+\]
+It will be convenient now to write $y_{1}$,~$u_{1}$ for $y$,~$u$, and to denote by
+\[
+y_{1},\ y_{2},\ \dots,\ y_{n},\quad
+u_{1},\ u_{2},\ \dots,\ u_{m},
+\]
+the complete systems of roots of the equations~\Eq{(1)}.
+
+We have
+\[
+\psi(x, u_{1}) = 0,
+\]
+and so
+\[
+\chi_{1}
+  = \frac{\dd\psi}{\dd x} + \frac{\dd\psi}{\dd u_{1}}\, \frac{du_{1}}{dx}
+  = \frac{\dd\psi}{\dd x} + y_{1} \frac{\dd\psi}{\dd u_{1}} = 0.
+\]
+Now let
+\[
+\Omega(x, u_{1})
+  = \prod_{r=1}^{n} \left(\frac{\dd\psi}{\dd x} + y_{r} \frac{\dd\psi}{\dd u_{1}}\right).
+\]
+Then $\Omega$~is a polynomial in~$u_{1}$, with coefficients symmetric in $y_{1}$, $y_{2}$,~\dots, $y_{n}$
+and therefore rational in~$x$.
+%% -----File: 049.png---
+
+The equations $\psi = 0$ and $\Omega = 0$ have a root~$u_{1}$ in common, and the
+first equation is irreducible. It follows, by Lemma~(2) of~\hyperlink{5 para 11.}{§11}, that
+\[
+\Omega(x, u_{s}) = 0
+\]
+for $s = 1$, $2$,~\dots, $m$.\footnote
+  {If $p(x)$~is the least common multiple of the denominators of the coefficients
+  of powers of~$u$ in~$\Omega$, then
+  \[
+  \Omega(x, u) p(x) = \chi(x, u),
+  \]
+  where $\chi$~is a polynomial. Applying Lemma~(2), we see that $\chi(x, u_{s}) = 0$, and so
+  \[
+  \Omega(x, u_{s}) = 0.
+  \]}
+And from this it follows that, when $s$~is given,
+we have
+\[
+\frac{\dd \psi}{\dd x} + y_{r} \frac{\dd \psi}{\dd u_{s}}
+\Tag{(2)}
+\]
+for some value of the suffix~$r$.
+
+But we have also
+\[
+\frac{\dd \psi}{\dd x} + \frac{\dd \psi}{\dd u_{s}}\, \frac{du_{s}}{dx} = 0;
+\Tag{(3)}
+\]
+and from \Eq{(2)}~and~\Eq{(3)} it follows\footnote
+  {It is impossible that $\psi$~and~$\dfrac{\dd \psi}{\dd u}$ should both vanish for $u = u_{s}$, since $\psi$~is
+  irreducible.}
+that
+\[
+\frac{du_{s}}{dx} = y_{r},
+\Tag{(4)}
+\]
+\textit{i.e.}\ that \emph{every~$u$ is the integral of some~$y$.}
+
+In the same way we can show that \emph{every~$y$ is the derivative of some~$u$.}
+Let
+\[
+\omega(x, y_{1})
+  = \prod_{s=1}^{m}\left(
+      \frac{\dd \psi}{\dd x} + y_{1}\frac{\dd \psi}{\dd u_{s}}
+  \right).
+\]
+Then $\omega$~is a polynomial in~$y_{1}$, with coefficients symmetric in $u_{1}$, $u_{2}$,~\dots, $u_{m}$
+and therefore rational in~$x$. The equations $f = 0$ and $\omega = 0$ have a
+root~$y_{1}$ in common, and so
+\[
+\omega(x, y_{r}) = 0
+\]
+for $r = 1$, $2$,~\dots, $n$. From this we deduce that, when $r$~is given, \Eq{(2)}~must
+be true for some value of~$s$, and so that the same is true of~\Eq{(4)}.
+
+Now it is impossible that, in~\Eq{(4)}, two different values of~$s$ should
+correspond to the same value of~$r$. For this would involve
+\[
+u_{s} - u_{t} = c
+\]
+where $s \neq t$ and $c$~is a constant. Hence we should have
+\[
+\psi(x, u_{s}) = 0,\quad
+\psi(x, u_{s} - c) = 0.
+\]
+%% -----File: 050.png---
+Subtracting these equations, we should obtain an equation of degree
+$m - 1$ in~$u_{s}$, with coefficients which are polynomials in~$x$; and this is
+impossible. In the same way we can prove that two different values of~$r$
+cannot correspond to the same value of~$s$.
+
+The equation~\Eq{(4)} therefore establishes a one-one correspondence
+between the values of $r$~and~$s$. It follows that
+\[
+m = n.
+\]
+It is moreover evident that, by arranging the suffixes properly, we can
+make
+\[
+\frac{du_{r}}{dx} = y_{r}
+\Tag{(5)}
+\]
+for $r = 1$, $2$,~\dots, $n$.
+
+\Paragraph{5}{13.} We have
+\[
+y_{r} = \frac{du_{r}}{dx}
+  = - \frac{\dd \psi}{\dd x} \bigg/ \frac{\dd \psi}{\dd u_{r}}
+  = R(x, u_{r}),
+\]
+where $R$~is a rational function which may, in virtue of Lemma~(3) of~\hyperlink{5 para 11.}{§~11},
+be expressed as a polynomial of degree $n - 1$ in~$u_{r}$, with coefficients
+rational in~$x$.
+
+The product
+\[
+\prod_{s \neq r} (z - y_{s})
+\]
+is a polynomial of degree $n - 1$ in~$z$, with coefficients which are symmetric
+polynomials in $y_{1}$, $y_{2}$,~\dots, $y_{r-1}$, $y_{r+1}$,~\dots,~$y_{n}$ and therefore,
+by \hyperlink{2 para 3.}{\textsc{ii.},~§3,~(ii)}, polynomials in~$y_{r}$ with coefficients rational in~$x$.
+Replacing $y_{r}$ by its expression as a polynomial in~$u_{r}$ obtained above,
+and eliminating $u_{r}^{n}$ and all higher powers of~$u_{r}$, we obtain an equation
+\[
+\prod_{s \neq r} (z - y_{s})
+  = \sum_{j=0}^{n-1} \sum_{k=0}^{n-1} S_{j,k}(x) z^{j} u_{r}^{k},
+\]
+where the $S$'s are rational functions of~$x$ which are, from the method
+of their formation, independent of the particular value of~$r$ selected.
+We may therefore write
+\[
+\prod_{s \neq r} (z - y_{s}) = P(x, z, u_{r}),
+\]
+where $P$~is a polynomial in $z$~and~$u_{r}$ with coefficients rational in~$x$. It
+is evident that
+\[
+P(x, y_{s}, u_{r}) = 0
+\]
+for every value of $s$ other than~$r$. In particular
+\[
+P(x, y_{1}, u_{r}) = 0\qquad
+(r = 2,\ 3,\ \dots,\DPtypo{}{\ n).}
+\]
+%% -----File: 051.png---
+
+It follows that the $n-1$~roots of the equation in~$u$
+\[
+P(x, y_{1}, u) = 0
+\]
+are $u_{2}$, $u_{3}$,~\dots, $u_{n}$. We have therefore
+\begin{align*}
+P(x, y_{1}, u)
+  &= T_{0}(x, y_{1}) \prod_{2}^{n} (u - u_{r})\\
+  &= T_{0}(x, y_{1}) \{u^{n-1} - u^{n-2}(u_{2} + u_{3} + \dots + u_{n}) + \dots\}\\
+  &= T_{0}(x, y_{1}) \left[u^{n-1} + u^{n-2}\left\{u_{1} + \frac{B_{1}(x)}{B_{0}(x)}\right\} + \dots\right],
+\end{align*}
+where $T_{0}(x, y_{1})$ is the coefficient of~$u^{n-1}$ in~$P$, and $B_{0}(x)$ and~$B_{1}(x)$
+are the coefficients of $u^{n}$~and~$u^{n-1}$ in~$\psi$. Equating the coefficients of~$u^{n-2}$
+on the two sides of this equation, we obtain
+\[
+u_{1} + \frac{B_{1}(x)}{B_{0}(x)} = \frac{T_{1}(x, y_{1})}{T_{0}(x, y_{1})},
+\]
+where $T_{1}(x, y_{1})$~is the coefficient of~$u^{n-2}$ in~$P$. Thus the theorem is
+proved.
+
+\Paragraph{5}{14.} We can now apply Lemma~(3) of §11; and we arrive at the
+final conclusion that \begin{Result}if
+\[
+\int y\,dx
+\]
+is algebraical then it can be expressed in the form
+\[
+R_{0} + R_{1}y + \dots + R_{n-1}y^{n-1},
+\]
+where $R_{0}$, $R_{1}$,~\dots\ are rational functions of~$x$.
+\end{Result}
+
+The most important case is that in which
+\[
+y = \sqrt[n]{R(x)},
+\]
+where $R(x)$~is rational. In this case
+\begin{align*}
+y^{n} &= R(x),
+\Tag{(1)}\\
+\frac{dy}{dx} &= \frac{R'(x)}{ny^{n-1}}.
+\Tag{(2)}
+\end{align*}
+
+But
+\begin{multline*}
+y = R_{0}' + R_{1}'y + \dots + R_{n-1}'y^{n-1} \\
+  + \{R_{1} + 2R_{2}y + \dots + (n - 1)R_{n-1}y^{n-2}\}\, \frac{dy}{dx}.
+\Tag{(3)}
+\end{multline*}
+Eliminating $\dfrac{dy}{dx}$ between these equations, we obtain an equation
+\[
+\varpi(x, y) = 0,
+\Tag{(4)}
+\]
+where $\varpi(x, y)$~is a polynomial. It follows from Lemma~(2) of §11
+that this equation must be satisfied by all the roots of~\Eq{(1)}. Thus~\Eq{(4)}
+is still true if we replace~$y$ by any other root~$y'$ of~\Eq{(1)}; and as
+%% -----File: 052.png---
+\Eq{(2)}~is still true when we effect this substitution, it follows that \Eq{(3)}~is
+also still true. Integrating, we see that the equation
+\[
+\int y\,dx = R_{0} + R_{1}y + \dots + R_{n-1}y^{n-1}
+\]
+is true when $y$~is replaced by~$y'$. We may therefore replace~$y$ by~$\omega y$,
+$\omega$~being any primitive $n$th~root of unity. Making this substitution,
+and multiplying by~$\omega^{n-1}$, we obtain
+\[
+\int y\,dx = \omega^{n-1}R_{0} + R_{1}y + \omega R_{2}y + \dots + \omega^{n-2}R_{n-1}y^{n-1};
+\]
+and on adding the $n$~equations of this type we obtain
+\[
+\int y\,dx = R_{1}y.
+\]
+Thus in this case the functions $R_{0}$, $R_{2}$,~\dots, $R_{n-1}$ all disappear.
+
+It has been shown by Liouville\footnote
+  {`Premier mémoire sur la détermination des intégrales dont la valeur est
+  algébrique', \textit{Journal de l'École Polytechnique}, vol.~14, cahier~22, 1833, pp.~124--148;
+  `Second mémoire\dots', \ibid., pp.~149--193.}
+that the preceding results enable
+us to obtain in all cases, by a finite number of elementary algebraical
+operations, a solution of the problem `\emph{to determine whether $\int y\,dx$ is
+algebraical, and to find the integral when it is algebraical}'.
+
+{\small \Paragraph{5}{15.} It would take too long to attempt to trace in detail the steps of the
+general argument. We shall confine ourselves to a solution of a particular
+problem which will give a sufficient illustration of the general nature of the
+arguments which must be employed.
+
+We shall determine under what circumstances the integral
+\[
+\int \frac{dx}{(x - p) \sqrt{ax^{2} + 2bx + c}}
+\]
+is algebraical. This question might of course be answered by actually
+evaluating the integral in the general case and finding when the integral
+function reduces to an algebraical function. We are now, however, in a
+position to answer it without any such integration.
+
+We shall suppose first that $ax^{2} + 2bx + c$ is not a perfect square. In this
+case
+\[
+y = \frac{1}{\sqrt{X}},
+\]
+where
+\[
+X = (x - p)^{2} (ax^{2} + 2bx + c),
+\]
+and if $\int y\,dx$ is algebraical it must be of the form
+\[
+\frac{R(x)}{\sqrt{X}}.
+\]
+Hence
+\[
+y = \frac{d}{dx}\left(\frac{R}{\sqrt{X}}\right),
+\]
+or
+\[
+2X = 2XR' - RX'.
+\]
+%% -----File: 053.png---
+
+We can now show that $R$~is a polynomial in~$x$. For if $R = U/V$, where $U$
+and~$V$ are polynomials, then~$V$, if not a mere constant, must contain a factor
+\[
+(x - \alpha)^{\mu}\qquad (\mu > 0),
+\]
+and we can put
+\[
+R = \frac{U}{W(x - \alpha)^{\mu}},
+\]
+where $U$~and~$W$ do not contain the factor~$x - \alpha$. Substituting this expression
+for~$R$, and reducing, we obtain
+\[
+\frac{2\mu UWX}{x - \alpha}
+  = 2U'WX - 2UW'X - UWX' - 2W^{2}X (x - \alpha)^{\mu}.
+\]
+Hence $X$~must be divisible by~$x - \alpha$. Suppose then that
+\[
+X = (x - \alpha)^{k} Y,
+\]
+where $Y$~is prime to~$x - \alpha$. Substituting in the equation last obtained we
+deduce
+\[
+\frac{(2\mu + k)UWY}{x - \alpha}
+  = 2U'WY - 2UW'Y - UWY' - 2W^{2}Y (x - \alpha)^{\mu},
+\]
+which is obviously impossible, since neither $U$,~$W$, nor~$Y$ is divisible by~$x - \alpha$.
+Thus $V$~must be a constant. Hence
+\[
+\int \frac{dx}{(x - p) \sqrt{ax^{2} + 2bx + c}}
+  = \frac{U(x)}{(x - p) \sqrt{ax^{2} + 2bx + c}},
+\]
+where $U(x)$~is a polynomial.
+
+Differentiating and clearing of radicals we obtain
+\[
+\{(x - p) (U' - 1) - U\} (ax^{2} + 2bx + c) = U(x - p)(ax + b).
+\]
+Suppose that the first term in~$U$ is~$Ax^{m}$. Equating the coefficients of~$x^{m+2}$,
+we find at once that $m = 2$. We may therefore take
+\[
+U = Ax^{2} + 2Bx + C,
+\]
+so that
+\begin{multline*}
+\{(x - p) (2Ax + 2B - 1) - Ax^{2} - 2Bx - C\} (ax^{2} + 2bx + c) \\
+  = (x - p) (ax + b) (Ax^{2} + 2Bx + C).
+\Tag{(1)}
+\end{multline*}
+
+From~\Eq{(1)} it follows that
+\[
+(x - p) (ax + b) (Ax^{2} + 2Bx + C)
+\]
+is divisible by $ax^{2} + 2bx + c$. But $ax + b$~is not a factor of $ax^{2} + 2bx + c$, as
+the latter is not a perfect square. Hence either (i)~$ax^{2} + 2bx + c$ and
+$Ax^{2} + 2Bx + C$ differ only by a constant factor or (ii)~the two quadratics have
+one and only one factor in common, and $x - p$~is also a factor of $ax^{2} + 2bx + c$.
+In the latter case we may write
+\[
+ax^{2} + 2bx + c = a(x - p) (x - q),\quad
+Ax^{2} + 2Bx + C = A(x - q) (x - r),
+\]
+where $p \neq q$, $p \neq r$. It then follows from~\Eq{(1)} that
+\[
+a(x - p) (2Ax + 2B - 1) - aA(x - q) (x - r) = A(ax + b) (x - r).
+\]
+Hence $2Ax + 2B - 1$ is divisible by~$x - r$. Dividing by~$aA(x - r)$ we obtain
+\[
+2(x - p) - (x - q) = x + \frac{b}{a} = x - \tfrac{1}{2}(p + q),
+\]
+and so $p = q$, which is untrue.
+%% -----File: 054.png---
+
+Hence case~(ii) is impossible, and so $ax^{2} + 2bx + c$ and $Ax^{2} + 2Bx + C$ differ
+only by a constant factor. It then follows from~\Eq{(1)} that $x - p$~is a factor
+of $ax^{2} + 2bx + c$; and the result becomes
+\[
+\int \frac{dx}{(x - p) \sqrt{ax^{2} + 2bx + c}}
+  = K\frac{\sqrt{ax^{2} + 2bx + c}}{x - p},
+\]
+where $K$~is a constant. It is easily verified that this equation is actually
+true when $ap^{2} + 2bp + c = 0$, and that
+\[
+K = \frac{1}{\sqrt{b^{2} - ac}}.
+\]
+The formula is equivalent to
+\[
+\int \frac{dx}{(x - p) \sqrt{(x - p)(x - q)}}
+  = \frac{2}{q - p} \sqrt{\frac{x - q}{x - p}}.
+\]
+
+There remains for consideration the case in which $ax^{2} + 2bx + c$ is a
+perfect square, say $a(x - q)^{2}$. Then
+\[
+\int \frac{dx}{(x - p)(x - q)}
+\]
+must be rational, and so $p = q$.
+
+As a further example, the reader may verify that if
+\[
+y^{3} - 3y + 2x = 0
+\]
+then
+\[
+\int y\,dx = \frac{3}{8} (2xy - y^{2}).\footnotemark
+\]
+  \footnotetext{Raffy, `Sur les quadratures algébriques et logarithmiques', \textit{Annales de l'École
+  Normale}, ser.~3, vol.~2, 1885, pp.~185--206.}}%
+
+\Paragraph{5}{16.} The theorem of §11 enables us to complete the proof of the
+two fundamental theorems stated without proof in~\hyperlink{2 para 5.}{\textsc{ii.},~§5}, viz.\
+
+\Item{(\ia)} $e^{x}$ is not an algebraical function of~$x$,
+
+\Item{(\ib)} $\log x$ is not an algebraical function of~$x$.
+
+We shall prove~(\ib) as a special case of a more general theorem, viz.\
+`\emph{no sum of the form
+\[
+A \log(x - \alpha) + B \log(x - \beta) + \dots,
+\]
+in which the coefficients $A$,~$B$,~\dots\ are not all zero, can be an algebraical
+function of~$x$}'. To prove this we have only to observe that the sum
+in question is the integral of a rational function of~$x$. If then it is
+algebraical it must, by the theorem of~§11, be rational, and this we
+have already seen to be impossible~(\hyperlink{4 para 2.}{\textsc{iv.},~2}).
+
+That $e^{x}$~is not algebraical now follows at once from the fact that it
+is the inverse function of~$\log x$.
+
+\Paragraph{5}{17.} The general theorem of~§11 gives the first step in the rigid
+proof of `Laplace's principle' stated in~\hyperlink{3 para 2.}{\textsc{iii.},~§2}. On account of the
+immense importance of this principle we repeat Laplace's words:
+%% -----File: 055.png---
+`\textit{l'intégrale d'une fonction différentielle ne peut contenir d'autres quantités
+\DPtypo{radicaux}{radicales} que celles qui entrent dans cette fonction}'. This general
+principle, combined with arguments similar to those used above~(\hyperlink{5 para 15.}{§15}) in
+a particular case, enables us to prove without difficulty that a great
+many integrals cannot be algebraical, notably the standard elliptic
+integrals
+\[
+\int \frac{dx}{\sqrt{(1 - x^{2}) (1 - k^{2}x^{2})}},\quad
+\int \sqrt{\frac{1 - x^{2}}{1 - k^{2}x^{2}}}\, dx,\quad
+\int \frac{dx}{\sqrt{4x^{3} - g_{2}x - g_{3}}}
+\]
+which give rise by inversion to the elliptic functions.
+
+\Paragraph{5}{18.} We must now consider in a very summary manner the more
+difficult question of the nature of those integrals of algebraical functions
+which are expressible in finite terms by means of the elementary
+transcendental functions. In the first place \emph{no integral of any algebraical
+function can contain any exponential}. Of this theorem it is, as
+we remarked before, easy to become convinced by a little reflection,
+as doubtless did Laplace, who certainly possessed no rigorous proof.
+The reader will find little difficulty in coming to the conclusion that
+exponentials cannot be eliminated from an elementary function by
+differentiation. But we would strongly recommend him to study the
+exceedingly beautiful and ingenious proof of this proposition given by
+Liouville\footnotemark.
+  \footnotetext{`Mémoire sur les transcendantes elliptiques considérées comme fonctions de
+  leur amplitude', \textit{Journal de l'École Polytechnique}, vol.~14, cahier~23, 1834,
+  pp.~37--83. The proof may also be found in Bertrand's \textit{Calcul intégral}, p.~99.}%
+We have unfortunately no space to insert it here.
+
+{\small It is instructive to consider particular cases of this theorem. Suppose for
+example that $\int y\,dx$, where $y$~is algebraical, were a polynomial in $x$~and~$e^{x}$, say
+\[
+\sum \sum a_{m, n}x^{m}e^{nx}.
+\Tag{(1)}
+\]
+When this expression is differentiated, $e^{x}$~must disappear from it: otherwise
+we should have an algebraical relation between $x$~and~$e^{x}$. Expressing the conditions
+that the coefficient of every power of~$e^{x}$ in the differential coefficient
+of~\Eq{(1)} vanishes identically, we find that the same must be true of~\Eq{(1)}, so that
+after all the integral does not really contain~$e^{x}$. Liouville's proof is in reality
+a development of this idea.}
+
+The integral of an algebraical function, if expressible in terms
+of elementary functions, can therefore only contain algebraical or
+logarithmic functions. The next step is to show that the logarithms
+must be simple logarithms of algebraical functions and can only
+enter linearly, so that the general integral must be of the type
+\[
+\int y\,dx = u + A \log v + B \log w + \dots,
+\]
+%% -----File: 056.png---
+where $A$,~$B$,~\dots\ are constants and $u$,~$v$, $w$,~\dots\ algebraical functions.
+Only when the logarithms occur in this simple form will differentiation
+eliminate them.
+
+Lastly it can be shown by arguments similar to those of \hyperlink{5 para 11.}{§§11--14}
+that $u$,~$v$, $w$,~\dots\ are rational functions of $x$~and~$y$. Thus $\int y\,dx$, if
+an elementary function, is \emph{the sum of a rational function of~$x$ and~$y$
+and of certain constant multiples of logarithms of such functions}.
+We can suppose that no two of $A$,~$B$,~\dots\ are commensurable, or indeed,
+more generally, that no linear relation
+\[
+A\alpha + B\beta + \dots = 0,
+\]
+with rational coefficients, holds between them. For if such a relation
+held then we could eliminate~$A$ from the integral, writing it in the
+form
+\[
+\int y\,dx = u + B \log(wv^{-\beta/\alpha}) + \dots\DPtypo{}{.}
+\]
+
+{\small It is instructive to verify the truth of this theorem in the special case in
+which the curve $f(x,y) = 0$ is unicursal. In this case $x$~and~$y$ are rational
+functions $R(t)$,~$S(t)$ of a parameter~$t$, and the integral, being the integral of
+a rational function of~$t$, is of the form
+\[
+u + A \log v + B \log w + \dots,
+\]
+where $u$,~$v$, $w$,~\dots\ are rational functions of~$t$. But $t$~may be expressed, by
+means of elementary algebraical operations, as a rational function of $x$~and~$y$.
+Thus $u$,~$v$, $w$,~\dots\ are rational functions of $x$~and~$y$.}
+
+The case of greatest interest is that in which $y$~is a rational function
+of $x$~and~$\sqrt{X}$, where $X$~is a polynomial. As we have already seen,
+$y$~can in this case be expressed in the form
+\[
+P + \frac{Q}{\sqrt{X}},
+\]
+where $P$~and~$Q$ are rational functions of~$x$. We shall suppress the
+rational part and suppose that $y = Q/\sqrt{X}$. In this case the general
+theorem gives
+\[
+\int\frac{Q}{\sqrt{X}}\,dx
+  = S + \frac{T}{\sqrt{X}}
+        + A \log(\alpha + \beta\sqrt{X})
+        + B \log(\gamma + \delta\sqrt{X}) + \dots,
+\]
+where $S$,~$T$, $\alpha$,~$\beta$, $\gamma$,~$\delta$,~\dots\ are rational. If we differentiate this equation
+we obtain an algebraical identity in which we can change the sign of~$\sqrt{X}$.
+Thus we may change the sign of~$\sqrt{X}$ in the integral equation.
+If we do this and subtract, and write $2A$,~\dots\ for $A$,~\dots, we obtain
+\[
+\int\frac{Q}{\sqrt{X}}\,dx
+  = \frac{T}{\sqrt{X}}
+    + A \log\frac{\alpha + \beta \sqrt{X}}{\alpha - \beta \sqrt{X}}
+    + B \log\frac{\gamma + \delta\sqrt{X}}{\gamma - \delta\sqrt{X}} + \dots,
+\]
+%% -----File: 057.png---
+which is the standard form for such an integral. It is evident that we
+may suppose $\alpha$,~$\beta$, $\gamma$,~\dots\ to be polynomials.
+
+{\small \Paragraph{5}{19.} \Item{(i)} By means of this theorem it is possible to prove that a number
+of important integrals, and notably the integrals
+\[
+\int \frac{dx}{\sqrt{(1 - x^{2}) (1 - k^{2}x^{2})}},\quad
+\int \sqrt{\frac{1 - x^{2}}{1 - k^{2}x^{2}}}\, dx,\quad
+\int \frac{dx}{\sqrt{4x^{3} - g_{2}x - g_{3}}},
+\]
+are not expressible in terms of elementary functions, and so represent genuinely
+new transcendents. The formal proof of this was worked out by Liouville\footnotemark;
+  \footnotetext{See Liouville's memoir quoted on \hyperlink{5 para 18.}{p.~45} (pp.~45~\textit{et~seq.}).}
+it rests merely on a consideration of the possible forms of the differential
+coefficients of expressions of the form
+\[
+\frac{T}{\sqrt{X}}
+  + A\log\frac{\alpha + \beta\sqrt{X}}
+              {\alpha - \beta\sqrt{X}} + \dots,
+\]
+and the arguments used are purely algebraical and of no great theoretical
+difficulty. The proof is however too detailed to be inserted here. It is not
+difficult to find shorter proofs, but these are of a less elementary character,
+being based on ideas drawn from the theory of functions\footnotemark.
+  \footnotetext{The proof given by Laurent (\textit{Traité d'analyse}, vol.~4, pp.~153~\textit{et~seq.}) appears at
+  first sight to combine the advantages of both methods of proof, but unfortunately
+  will not bear a closer examination.}
+
+The general questions of this nature which arise in connection with
+integrals of the form
+\[
+\int \frac{Q}{\sqrt{X}}\, dx,
+\]
+or, more generally,
+\[
+\int \frac{Q}{\sqrt[m]{X}}\, dx,
+\]
+are of extreme interest and difficulty. The case which has received most
+attention is that in which $m = 2$ and $X$~is of the third or fourth degree, in
+which case the integral is said to be \emph{elliptic}. An integral of this kind is
+called \emph{pseudo-elliptic} if it is expressible in terms of algebraical and logarithmic
+functions. Two examples were given above (\hyperlink{5 para 10.}{§~10}). General methods have
+been given for the construction of such integrals, and it has been shown that
+certain interesting forms are pseudo-elliptic. In Goursat's \textit{Cours d'analyse}\footnotemark,
+  \footnotetext{Second edition, vol.~1, pp.~267--269.}%
+for instance, it is shown that if $f(x)$~is a rational function such that
+\[
+f(x) + f\left(\frac{1}{k^{2}x}\right) = 0,
+\]
+then
+\[
+\int \frac{f(x)\, dx}{\sqrt{x(1 - x) (1 - k^{2}x)}}
+\]
+is pseudo-elliptic. But no method has been devised as yet by which we can
+always determine in a finite number of steps whether a \emph{given} elliptic integral
+%% -----File: 058.png---
+is pseudo-elliptic, and integrate it if it is, and there is reason to suppose that
+no such method can be given. And up to the present it has not, so far as
+we know, been proved rigorously and explicitly that (\textit{e.g.})\ the function
+\[
+u = \int \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}
+\]
+is not a root of an elementary transcendental equation; all that has been
+shown is that it is not \emph{explicitly} expressible in terms of elementary transcendents.
+The processes of reasoning employed here, and in the memoirs
+to which we have referred, do not therefore suffice to prove that the inverse
+function $x = \sn u$ is not an elementary function of~$u$. Such a proof must rest
+on the known properties of the function $\sn u$, and would lie altogether outside
+the province of this tract.
+
+The reader who desires to pursue the subject further will find references
+to the original authorities in \hyperlink{Appendix I}{Appendix~I}.
+
+\Item{(ii)} One particular class of integrals which is of especial interest is
+that of the \emph{binomial integrals}
+\[
+\int x^{m}(ax^{n} + b)^{p}\, dx,
+\]
+where $m$,~$n$,~$p$ are rational. Putting $ax^{n} = bt$, and neglecting a constant
+factor, we obtain an integral of the form
+\[
+\int t^{q}(1 + t)^{p}\, dt,
+\]
+where $p$~and~$q$ are rational. If $p$~is an integer, and $q$~a fraction~$r/s$, this
+integral can be evaluated at once by putting $t = u^{s}$, a substitution which
+rationalises the integrand. If $q$~is an integer, and $p = r/s$, we put $1 + t = u^{s}$.
+If $p + q$ is an integer, and $p = r/s$, we put $1 + t = tu^{s}$.
+
+It follows from Tschebyschef's researches (to which references are given
+in Appendix~I) that these three cases are the only ones in which the integral
+can be evaluated in finite form.}
+
+\Paragraph{5}{20.} In \hyperlink{5 para 7.}{§§7--9} we considered in some detail the integrals connected
+with curves whose deficiency is zero. We shall now consider
+in a more summary way the case next in simplicity, that in which
+the deficiency is unity, so that the number of double points is
+\[
+\tfrac{1}{2} (n - 1)(n - 2) - 1 = \tfrac{1}{2} n(n - 3).
+\]
+It has been shown by Clebsch\footnote
+  {`Über diejenigen Curven, deren Coordinaten sich als elliptische Functionen
+  eines Parameters darstellen lassen', \textit{Journal für Mathematik}, vol.~64, 1865,
+  pp.~210--270.}
+that in this case the coordinates of
+the points of the curve can be expressed as \emph{rational functions of
+a parameter~$t$ and of the square root of a polynomial in~$t$ of the third
+or fourth degree}.
+%% -----File: 059.png---
+
+{\small The fact is that the curves
+\begin{align*}
+y^{2} &= a + bx + cx^{2} + dx^{3}, \\
+y^{2} &= a + bx + cx^{2} + dx^{3} + ex^{4},
+\end{align*}
+are the simplest curves of deficiency~$1$. The first is the typical cubic
+without a double point. The second is a quartic with two double points,
+in this case coinciding in a `tacnode' at infinity, as we see by making the
+equation homogeneous with~$z$, writing $1$ for~$y$, and then comparing the
+resulting equation with the form treated by Salmon on p.~215 of his \textit{Higher
+plane curves}. The reader who is familiar with the theory of algebraical plane
+curves will remember that the deficiency of a curve is unaltered by any
+birational transformation of coordinates, and that any curve can be birationally
+transformed into any other curve of the same deficiency, so that any
+curve of deficiency~$1$ can be birationally transformed into the cubic whose
+equation is written above.}
+
+The argument by which this general theorem is proved is very
+much like that by which we proved the corresponding theorem for
+unicursal curves. The simplest case is that of the general cubic curve.
+We take a point on the curve as origin, so that the equation of the
+curve is of the form
+\[
+ax^{3} + 3bx^{2}y + 3cxy^{2} + dy^{3}
+  + ex^{2} + 2fxy + gy^{2}
+  + hx + ky = 0.
+\]
+Let us consider the intersections of this curve with the secant $y = tx$.
+Eliminating~$y$, and solving the resulting quadratic in~$x$, we see that the
+only irrationality which enters into the expression of~$x$ is
+\[
+\sqrt {T_{2}^{2} - 4T_{1}T_{3}},
+\]
+where
+\[
+T_{1} = h + kt,\quad
+T_{2} = e + 2ft + gt^{2},\quad
+T_{3} = a + 3bt + 3ct^{2} + dt^{3}.
+\]
+
+A more elegant method has been given by Clebsch\footnotemark.
+  \footnotetext{See Hermite, \textit{Cours d'analyse}, pp.~422--425.}
+If we
+write the cubic in the form
+\[
+LMN = P,
+\]
+where $L$,~$M$,~$N$, $P$ are linear functions of $x$~and~$y$, so that $L$,~$M$,~$N$ are
+the asymptotes, then the hyperbolas $LM = t$ will meet the cubic in
+four fixed points at infinity, and therefore in two points only which
+depend on~$t$. For these points
+\[
+LM = t,\quad P = tN.
+\]
+Eliminating~$y$ from these equations, we obtain an equation of the form
+\[
+Ax^{2} + 2Bx + C = 0,
+\]
+where $A$,~$B$,~$C$ are quadratics in~$t$. Hence
+\[
+x = -\frac{B}{A} ± \frac{\sqrt{B^{2} - AC}}{A} = R(t, \sqrt{T}),
+\]
+%% -----File: 060.png---
+where $T = B^{2} - AC$ is a polynomial in~$t$ of degree not higher than the
+fourth.
+
+Thus if the curve is
+\[
+x^{3} + y^{3} - 3axy + 1 = 0,
+\]
+so that
+\[
+L = \omega x + \omega^{2} y + a, \quad
+M = \omega^{2} x + \omega y + a, \quad
+N = x + y + a, \quad
+P = a^{3} - 1,
+\]
+$\omega$~being an imaginary cube root of unity, then we find that the line
+\[
+x + y + a = \frac{a^{3} - 1}{t}
+\]
+meets the curve in the points given by
+\[
+x = \frac{b - at}{2t} ± \frac{\sqrt{3T}}{6t}, \quad
+y = \frac{b - at}{2t} \mp \frac{\sqrt{3T}}{6t},
+\]
+where $b = a^{3} - 1$ and
+\[
+T = 4t^{3} - 9a^{2}t^{2} + 6abt - b^{2}.
+\]
+
+In particular, for the curve
+\[
+x^{3} + y^{3} + 1 = 0,
+\]
+we have
+\[
+x = \frac{-\sqrt{3} + \sqrt{4t^{3} - 1}}{2t \sqrt{3}}, \quad
+y = \frac{-\sqrt{3} - \sqrt{4t^{3} - 1}}{2t \sqrt{3}}.
+\]
+
+\Paragraph{5}{21.} It will be plain from what precedes that
+\[
+\int R\{x, \sqrt[3]{a + bx + cx^{2} + dx^{3}}\} \,dx
+\]
+can always be reduced to an elliptic integral, the deficiency of the cubic
+\[
+y^{3} = a + bx + cx^{2} + dx^{3}
+\]
+being unity.
+
+In general integrals associated with curves whose deficiency is
+greater than unity cannot be so reduced. But associated with every
+curve of, let us say, deficiency~$2$ there will be an infinity of integrals
+\[
+\int R(x, y) \,dx
+\]
+reducible to elliptic integrals or even to elementary functions; and
+there are curves of deficiency~$2$ for which \emph{all} such integrals are
+reducible.
+
+For example, the integral
+\[
+\int R\{x, \sqrt{x^{6} + ax^{4} + bx^{2} + c}\} \,dx
+\]
+%% -----File: 061.png---
+may be split up into the sum of the integral of a rational function and
+two integrals of the types
+\[
+\int \frac{R(x^{2})\,dx}{\sqrt{x^{6} + ax^{4} + bx^{2} + c}},\
+\int \frac{xR(x^{2})\,dx}{\sqrt{x^{6} + ax^{4} + bx^{2} + c}},
+\]
+and each of these integrals becomes elliptic on putting $x^{2} = t$. But
+the deficiency of
+\[
+y^{2} = x^{6} + ax^{4} + bx^{2} + c
+\]
+is~$2$. Another example is given by the integral
+\[
+\int R\{x,\sqrt[4]{x^{4} + ax^{3} + bx^{2} + cx + d}\}\,dx.\footnotemark
+\]
+\footnotetext{See Legendre, \textit{Traité des fonctions elliptiques}, vol.~1, chs.~26--27, 32--33;
+Bertrand, \textit{Calcul intégral}, pp.~67 \textit{et seq.}; and Enneper, \textit{Elliptische Funktionen},
+note~1, where abundant references are given.}
+
+\Paragraph{5}{22.} It would be beside our present purpose to enter into any
+details as to the general theory of elliptic integrals, still less of the
+integrals (usually called Abelian) associated with curves of deficiency
+greater than unity. We have seen that if the deficiency is unity then
+the integral can be transformed into the form
+\[
+\int R(x, \sqrt{X})\,dx
+\]
+where
+\[
+X = x^{4} + ax^{3} + bx^{2} + cx + d.\footnotemark
+\]
+\footnotetext{There is a similar theory for curves of deficiency 2, in which $X$ is of the \emph{sixth}
+degree.}%
+It can be shown that, by a transformation of the type
+\[
+x =\frac{\alpha t + \beta}{\gamma t + \delta},
+\]
+this integral can be transformed into an integral
+\[
+\int R(t, \sqrt{T})\,dt
+\]
+where
+\[
+ T = t^{4} + At^{2} + B.
+\]
+
+We can then, as when $T$~is of the second degree (§3), decompose
+this integral into two integrals of the forms
+\[
+\int R(t)\,dt,\ \int \frac{R(t)\,dt}{\sqrt{T}}.
+\]
+Of these integrals the first is elementary, and the second can be
+%% -----File: 062.png---
+decomposed\footnote
+  {See, \textit{e.g.}, Goursat, \textit{Cours d'analyse}, ed.~2, vol.~1, pp.~257~\textit{et~seq.}}
+into the sum of an algebraical term, of certain multiples
+of the integrals
+\[
+\int\frac{dt}{\sqrt{T}},\quad
+\int\frac{t^{2}\,dt}{\sqrt{T}},
+\]
+and of a number of integrals of the type
+\[
+\int \frac{dt}{(t - \tau)\sqrt{T}}.
+\]
+These integrals cannot in general be reduced to elementary functions,
+and are therefore new transcendents.
+
+We will only add, before leaving this part of our subject, that the
+algebraical part of these integrals can be found by means of the
+elementary algebraical operations, as was the case with the rational
+part of the integral of a rational function, and with the algebraical part
+of the simple integrals considered in \hyperlink{5 para 14.}{§§14--15}.
+
+\Section{VI. Transcendental functions}
+
+\Paragraph{6}{1.} The theory of the integration of transcendental functions is
+naturally much less complete than that of the integration of rational
+or even of algebraical functions. It is obvious from the nature of the
+case that this must be so, as there is no general theorem concerning
+transcendental functions which in any way corresponds to the theorem
+that any algebraical combination of algebraical functions may be
+regarded as a simple algebraical function, the root of an equation of
+a simple standard type.
+
+It is indeed almost true to say that there is no general theory, or
+that the theory reduces to an enumeration of the few cases in which
+the integral may be transformed by an appropriate substitution into an
+integral of a rational or algebraical function. These few cases are
+however of great importance in applications.
+
+\Paragraph{6}{2.} \Item{(i)} The integral
+\[
+\int F(e^{ax}, e^{bx}, \dots, e^{kx})\,dx
+\]
+where $F$~is an algebraical function, and $a$,~$b$,~\dots, $k$ commensurable
+numbers, can always be reduced to that of an algebraical function.
+In particular the integral
+\[
+\int R(e^{ax}, e^{bx}, \dots, e^{kx})\,dx,
+\]
+%% -----File: 063.png---
+where $R$~is rational, is always an elementary function. In the first
+place a substitution of the type $x = \alpha y$ will reduce it to the form
+\[
+\int R(e^{y})\,dy,
+\]
+and then the substitution $e^{y} = z$ will reduce this integral to the integral
+of a rational function.
+
+{\Loosen In particular, since  $\cosh x$ and $\sinh x$ are rational functions of~$e^{x}$,
+and $\cos x$~and~$\sin x$ are rational functions of~$e^{ix}$, the integrals}
+\[
+\int R(\cosh x, \sinh x)\,dx,\quad
+\int R(\cos x, \sin x)\,dx
+\]
+are always elementary functions. In the second place the substitution
+just indicated is imaginary, and it is generally more convenient
+to use the substitution
+\[
+\tan \tfrac{1}{2} x = t,
+\]
+which reduces the integral to that of a rational function, since
+\[
+\cos x = \frac{1 - t^{2}}{1 + t^{2}},\quad
+\sin x = \frac{2t}{1 + t^{2}},\quad
+dx = \frac{2\, dt}{1 + t^{2}}.
+\]
+
+\Item{(ii)} The integrals
+\begin{gather*}
+\int R(\cosh x, \sinh x, \cosh 2x,\dots \sinh mx)\,dx, \\
+\int R(\cos x, \sin x, \cos 2x,\dots \sin mx)\,dx,
+\end{gather*}
+are included in the two standard integrals above.
+
+Let us consider some further developments concerning the integral
+\[
+\int R(\cos x, \sin x)\,dx.\footnotemark
+\]
+\footnotetext{See Hermite, \textit{Cours d'analyse}, pp.~320~\textit{et~seq}.}%
+If we make the substitution $z = e^{ix}$, the subject of integration becomes a
+rational function~$H(z)$, which we may suppose split up into
+
+\Item{(\ia)} a constant and certain positive and negative powers of~$z$,
+
+\Item{(\ib)} groups of terms of the type
+\[
+\frac{A_{0}}{z - a}
+  + \frac{A_{1}}{(z - a)^{2}} + \dots
+  + \frac{A_{n}}{(z - a)^{n+1}}.
+\Tag{(1)}
+\]
+
+The terms~(i), when expressed in terms of~$x$, give rise to a term
+\[
+\sum (c_{k}\cos kx + d_{k}\sin kx).
+\]
+In the group~\Eq{(1)} we put $z = e ^{ix}$, $a = e^{i\alpha}$ and, using the equation
+\[
+\frac{1}{z - a}
+  = \tfrac{1}{2} e^{-i\alpha} \{-1 - i \cot \tfrac{1}{2} (x - \alpha)\},
+\]
+%% -----File: 064.png---
+we obtain a polynomial of degree~$n + 1$ in $\cot \frac{1}{2} (x - \alpha)$. Since
+\[
+\cot^{2} x = -1 - \frac{d \cot x}{dx},\quad
+\cot^{3} x = -\cot x - \frac{1}{2}\, \frac{d}{dx} (\cot^{2} x),\ \dots,
+\]
+this polynomial may be transformed into the form
+\[
+C + C_{0} \cot \tfrac{1}{2} (x - \alpha)
+  + C_{1} \frac{d}{dx} \cot \tfrac{1}{2} (x - \alpha) + \dots
+  + C_{n} \frac{d^{n}}{dx^{n}} \cot \tfrac{1}{2} (x - \alpha).
+\]
+
+The function $R(\cos x, \sin x)$ is now expressed as a sum of a number of
+terms each of which is immediately integrable. The integral is a rational
+function of $\cos x$~and~$\sin x$ if all the constants~$C_{0}$ vanish; otherwise it includes
+a number of terms of the type
+\[
+2C_{0} \log \sin \tfrac{1}{2} (x - \alpha).
+\]
+
+Let us suppose for simplicity that~$H(z)$, when split up into partial fractions,
+contains no terms of the types
+\[
+C,\quad z^{m},\quad z^{-m},\quad (z - a)^{-p}\qquad (p > 1).
+\]
+Then
+\[
+R(\cos x, \sin x)
+  = C_{0} \cot \tfrac{1}{2} (x - \alpha)
+  + D_{0} \cot \tfrac{1}{2} (x - \beta) + \dots,
+\]
+and the constants $C_{0}$, $D_{0}$,~\dots\ may be determined by multiplying each side of
+the equation by $\sin \frac{1}{2} (x - \alpha)$, $\sin \frac{1}{2} (x - \beta)$,~\dots\ and making $x$~tend to $\alpha$,~$\beta$,~\dots.
+
+It is often convenient to use the equation
+\[
+\cot \tfrac{1}{2} (x - \alpha)
+  = \cot(x - \alpha) + \cosec(x - \alpha)
+\]
+which enables us to decompose the function~$R$ into two parts $U(x)$~and~$V(x)$
+such that
+\[
+U(x + \pi) = U(x),\quad
+V(x + \pi) = -V(x).
+\]
+If $R$~has the period~$\pi$, then $V$~must vanish identically; if it changes sign
+when $x$~is increased by~$\pi$, then $U$~must vanish identically. Thus we find
+without difficulty that, if $m < n$,
+\[
+\frac{\sin mx}{\sin nx}
+  = \frac{1}{2n} \sum_{0}^{2n-1}\frac{(-1)^{k}\sin m\alpha}{\sin(x - \alpha)}
+  = \frac{1}{n} \sum_{0}^{n-1}  \frac{(-1)^{k}\sin m\alpha}{\sin(x - \alpha)},
+\]
+or
+\[
+\frac{\sin mx}{\sin nx}
+  = \frac{1}{n} \sum_{0}^{n-1} (-1)^{k}\sin m\alpha \cot(x - \alpha),
+\]
+where $\alpha = k\pi/n$, according as $m + n$~is odd or even.
+
+Similarly
+\begin{gather*}
+\frac{1}{\sin(x - a)\sin(x - b)\sin(x - c)}
+  = \sum\frac{1}{\sin(a - b)\sin(a - c)\sin(x - a)},\\
+\frac{\sin(x - d)}{\sin(x - a)\sin(x - b)\sin(x - c)}
+  = \sum \frac{\sin(a - d)}{\sin(a - b)\sin(a - c)} \cot(x - a).
+\end{gather*}
+
+\Item{(iii)} One of the most important integrals in applications is
+\[
+\int \frac{dx}{a + b \cos x},
+\]
+where $a$~and~$b$ are real. This integral may be evaluated in the manner
+explained above, or by the transformation $\tan \frac{1}{2} x = t$. A more elegant method
+%% -----File: 065.png---
+is the following. If $|a| > |b|$, we suppose $a$~positive, and use the transformation
+\[
+(a + b\cos x)(a - b\cos y) = a^{2} - b^{2},
+\]
+which leads to
+\[
+\frac{dx}{a + b\cos x} = \frac{dy}{\sqrt{a^{2} - b^{2}}}.
+\]
+If $|a| < |b|$, we suppose $b$~positive, and use the transformation
+\[
+(b\cos x + a)(b\cosh y - a) = b^{2} - a^{2}.
+\]
+
+The integral
+\[
+\int \frac{dx}{a + b\cos x + c\sin x}
+\]
+may be reduced to this form by the substitution $x + a = y$, where $\cot a = b/c$.
+The forms of the integrals
+\[
+\int \frac{dx}{(a + b\cos x)^{n}},\quad
+\int \frac{dx}{(a + b\cos x + c\sin x)^{n}}
+\]
+may be deduced by the use of formulae of reduction, or by differentiation
+with respect to~$a$. The integral
+\[
+\int \frac{dx}{(A\cos^{2}x + 2B\cos x\sin x + C\sin^{2}x)^{n}}
+\]
+is really of the same type, since
+\[
+A\cos^{2}x + 2B\cos x\sin x + C\sin^{2}x
+  = \tfrac{1}{2} (A + C) + \tfrac{1}{2} (A - C)\cos 2x + B\sin 2x.
+\]
+And similar methods may be applied to the corresponding integrals which
+contain hyperbolic functions, so that this type includes a large variety of
+integrals of common occurrence.
+
+\Item{(iv)} The same substitutions may of course be used when the subject of
+integration is an irrational function of $\cos x$~and~$\sin x$, though sometimes
+it is better to use the substitutions $\cos x = t$, $\sin x = t$, or $\tan x = t$. Thus
+the integral
+\[
+\int R(\cos x, \sin x, \sqrt{X})\,dx,
+\]
+where
+\[
+X = (a, b, c, f, g, h\between \cos x, \sin x, 1)^{2},
+\]
+is reduced to an elliptic integral by the substitution $\tan \frac{1}{2} x = t$. The most
+important integrals of this type are
+\[
+\int \frac{R(\cos x, \sin x)\,dx}{\sqrt{1 - k^{2}\sin^{2}x}},\quad
+\int \frac{R(\cos x, \sin x)\,dx}{\sqrt{\alpha + \beta\cos x + \gamma\sin x}}.
+\]
+
+\Paragraph{6}{3.} The integral
+\[
+\int P(x, e^{ax}, e^{bx}, \dots, e^{kx})\,dx,
+\]
+where $a$, $b$,~\dots, $k$ are any numbers (commensurable or not), and $P$~is
+a polynomial, is always an elementary function. For it is obvious
+%% -----File: 066.png---
+that the integral can be reduced to the sum of a finite number of
+integrals of the type
+\[
+\int x^{p}e^{Ax}\,dx;
+\]
+and
+\[
+\int x^{p}e^{Ax}\,dx
+  = \left(\frac{\dd}{\dd A}\right)^{p} \int e^{Ax}\,dx
+  = \left(\frac{\dd}{\dd A}\right)^{p} \frac{e^{Ax}}{A}.
+\]
+This type of integral includes a large variety of integrals, such as
+\begin{align*}
+\int x^{m}(\cos px)^{\mu}(\sin qx)^{\nu}\,dx,\quad &
+\int x^{m}(\cosh px)^{\mu}(\sinh qx)^{\nu}\,dx,\\
+%
+\int x^{m}e^{-\alpha x}(\cos px)^{\mu}\,dx,\quad &
+\int x^{m}e^{-\alpha x}(\sin qx)^{\nu}\,dx,
+\end{align*}
+($m$,~$\mu$,~$\nu$, being positive integers) for which formulae of reduction are
+given in text-books on the integral calculus.
+
+Such integrals as
+\[
+\int P(x, \log x)\,dx,\quad
+\int P(x, \arcsin x)\,dx,\ \dots,
+\]
+where $P$~is a polynomial, may be reduced to particular cases of the
+above general integral by the obvious substitutions
+\[
+x = e^{y},\quad x = \sin y,\ \dots.
+\]
+
+\Paragraph{6}{4.} Except for the two classes of functions considered in the three
+preceding paragraphs, there are no really general classes of transcendental
+functions which we can \emph{always} integrate in finite terms, although
+of course there are innumerable particular forms which may be
+integrated by particular devices. There are however many classes
+of such integrals for which a systematic reduction theory may be given,
+analogous to the reduction theory for elliptic integrals. Such a reduction
+theory endeavours in each case
+
+\Item{(i)} to split up any integral of the class under consideration into
+the sum of a number of parts of which some are elementary and
+the others not;
+
+\Item{(ii)} to reduce the number of the latter terms to the least possible;
+
+\Item{(iii)} to prove that these terms are incapable of further reduction,
+and are genuinely new and independent transcendents.
+
+As an example of this process we shall consider the integral
+\[
+\int e^{x}R(x)\,dx
+\]
+where $R(x)$~is a rational function of~$x$.\footnote
+  {See Hermite, \textit{Cours d'analyse}, pp.~352~\textit{et~seq.}}
+The theory of partial
+%% -----File: 067.png---
+fractions enables us to decompose this integral into the sum of a
+number of terms
+\[
+A \int\frac{e^{x}}{x - a}\,dx,\DPtypo{}{\ \dots,}\quad
+A_{m} \int\frac{e^{x}}{(x - a)^{m+1}}\,dx,\DPtypo{\ \dots,}{}\quad
+B \int\frac{e^{x}}{x - b}\,dx,\ \dots.
+\]
+
+Since
+\[
+\int\frac{e^{x}}{(x - a)^{m+1}}\,dx
+  = -\frac{e^{x}}{m(x - a)^{m}}
+   + \frac{1}{m} \int\frac{e^{x}}{(x - a)^{m}}\,dx,
+\]
+the integral may be further reduced so as to contain only
+
+\Item{(i)} a term
+\[
+e^{x}S(x)
+\]
+where $S(x)$~is a rational function;
+
+\Item{(ii)} a number of terms of the type
+\[
+\alpha \int\frac{e^{x}\,dx}{x - a}.
+\]
+If all the constants~$\alpha$ vanish, then the integral can be calculated in the
+finite form~$e^{x}S(x)$. If they do not we can at any rate assert that the
+integral cannot be calculated \emph{in this form}\footnotemark.
+  \footnotetext{See the remarks at the end of this paragraph.}%
+For no such relation as
+\[
+\alpha \int\frac{e^{x}\,dx}{x - a}
+  + \beta  \int\frac{e^{x}\,dx}{x - b} + \dots
+  + \kappa \int\frac{e^{x}\,dx}{x - k}
+  = e^{x}T(x),
+\]
+where $T$~is rational, can hold for all values of~$x$. To see this it is
+only necessary to put $x = a + h$ and to expand in ascending powers
+of~$h$. Then
+\begin{align*}
+\alpha \int\frac{e^{x}\,dx}{x - a}
+  &= \alpha e^{a} \int\frac{e^{h}}{h}\,dh\\
+  &= \alpha e^{a} (\log h + h + \dots),
+\end{align*}
+and no \emph{logarithm} can occur in any of the other terms\footnotemark.
+  \footnotetext{It is not difficult to give a purely algebraical proof on the lines of \hyperlink{4 para 2.}{\textsc{iv}.,~§2}.}
+
+Consider, for example, the integral
+\[
+\int e^{x}\left(1 - \frac{1}{x}\right)^{3} dx.
+\]
+This is equal to
+\[
+e^{x} - 3 \int \frac{e^{x}}{x}\,dx
+      + 3 \int \frac{e^{x}}{x^{2}}\,dx
+      -   \int \frac{e^{x}}{x^{3}}\,dx,
+\]
+and since
+\[
+3 \int \frac{e^{x}}{x^{2}}\,dx
+  = -\frac{3e^{x}}{x} + 3 \int \frac{e^{x}}{x}\,dx,
+\]
+and
+\[
+-\int \frac{e^{x}}{x^{3}}\,dx
+  = \frac{e^{x}}{2x^{2}} - \frac{1}{2} \int \frac{e^{x}}{x^{2}}\,dx
+  = \frac{e^{x}}{2x^{2}} + \frac{e^{x}}{2x} - \frac{1}{2} \int \frac{e^{x}}{x}\,dx,
+\]
+%% -----File: 068.png---
+we obtain finally
+\[
+\int e^{x}\left(1 - \frac{1}{x}\right)^{3} dx
+   = e^{x}\left(1 - \frac{7}{2x} + \frac{1}{2x^{2}}\right)
+     - \frac{1}{2} \int \frac{e^{x}}{x}\, dx.
+\]
+Similarly it will be found that
+\[
+\int e^{x}\left(1 - \frac{2}{x}\right)^{2} dx
+  = 2e^{x}\left(\frac{1}{2} - \frac{2}{x}\right),
+\]
+this integral being an elementary function.
+
+Since
+\[
+\int \frac{e^{x}}{x - a}\, dx = e^{a} \int \frac{e^{y}}{y}\, dy,
+\]
+if $x = y + a$, all integrals of this kind may be made to depend on known
+functions and on the single transcendent
+\[
+\int \frac{e^{x}}{x}\, dx,
+\]
+which is usually denoted by $\Li e^{x}$ and is of great importance in the
+theory of numbers. The question of course arises as to whether this
+integral is not itself an elementary function.
+
+Now Liouville\footnote
+  {`Mémoire sur l'intégration d'une classe de fonctions transcendantes', \textit{Journal
+  für Mathematik}, vol.~13, 1835, pp.~93--118. Liouville shows how the integral, when
+  of this form, may always be calculated by elementary methods.}
+has proved the following theorem: \begin{Result}`if $y$~is any
+algebraical function of~$x$, and
+\[
+\int e^{x}y\,dx
+\]
+is an elementary function, then
+\[
+\int e^{x}y\,dx = e^{x}(\alpha + \beta y + \dots + \lambda y^{n-1}),
+\]
+$\alpha$, $\beta$,~\dots, $\lambda$ being rational functions of~$x$ and $n$~the degree of the
+algebraical equation which determines~$y$ as a function of~$x$'.
+\end{Result}
+
+Liouville's proof rests on the same general principles as do those of
+the corresponding theorems concerning the integral $\int y\,dx$. It will
+be observed that no logarithmic terms can occur, and that the theorem
+is therefore very similar to that which holds for $\int y\,dx$ in the simple
+case in which the integral is \emph{algebraical}. The argument which shows
+that no logarithmic terms occur is substantially the same as that which
+shows that, when they occur in the integral of an algebraical function,
+they must occur linearly. In this case the occurrence of the exponential
+factor precludes even this possibility, since differentiation
+will not eliminate logarithms when they occur in the form
+\[
+e^{x} \log f(x).
+\]
+%% -----File: 069.png---
+
+In particular, if $y$~is a rational function, then the integral must
+be of the form
+\[
+e^{x} R(x)
+\]
+and this we have already seen to be impossible. Hence the `logarithm-integral'
+\[
+\Li e^{x} = \int \frac{e^{x}}{x}\, dx = \int^{e^{x}} \frac{dy}{\log y}
+\]
+is really a new transcendent, which cannot be expressed in finite terms
+by means of elementary functions; and the same is true of all integrals
+of the type
+\[
+\int e^{x} R(x)\,dx
+\]
+which cannot be calculated in finite terms by means of the process of
+reduction sketched above.
+
+The integrals
+\[
+\int \sin x R(x)\,dx,\quad
+\int \cos x R(x)\,dx
+\]
+may be treated in a similar manner. Either the integral is of the form
+\[
+\cos x R_{1}(x) + \sin x R_{2}(x)
+\]
+or it consists of a term of this kind together with a number of terms
+which involve the transcendents
+\[
+\int \frac{\cos x}{x}\,dx,\quad
+\int \frac{\sin x}{x}\,dx,
+\]
+which are called the cosine-integral and sine-integral of~$x$, and denoted
+by $\Ci x$ and $\Si x$. These transcendents are of course not fundamentally
+distinct from the logarithm-integral.
+
+\Paragraph{6}{5.} Liouville has gone further and shown that it is always possible
+to determine whether the integral
+\[
+\int (Pe^{p} + Qe^{q} + \dots + Te^{t})\,dx,
+\]
+where $P$, $Q$,~\dots, $T$, $p$, $q$,~\dots, $t$ are algebraical functions, is an elementary
+function, and to obtain the integral in case it is one\footnotemark.
+  \footnotetext{An interesting particular result is that the `error function' $\int e^{-x^{2}}\,dx$ is not an
+  elementary function.}%
+The most
+general theorem which has been proved in this region of mathematics,
+and which is also due to Liouville, is the following.
+%% -----File: 070.png---
+
+\begin{Result}
+`If $y$, $z$,~\dots\ are functions of~$x$ whose differential coefficients are
+algebraical functions of $x$,~$y$, $z$,~\dots, and $F$~denotes an algebraical
+function, and if
+\[
+\int F(x, y, z, \dots)\,dx
+\]
+is an elementary function, then it is of the form
+\[
+t + A \log u + B \log v + \dots,
+\]
+where $t$, $u$, $v$,~\dots\ are algebraical functions of $x$,~$y$, $z$,~\dots. If the
+differential coefficients are rational in $x$,~$y$, $z$,~\dots, and $F$~is rational,
+then $t$,~$u$, $v$,~\dots\ are rational in $x$,~$y$, $z$,~\dots.'
+\end{Result}
+
+Thus for example the theorem applies to
+\[
+F(x, e^{x}, e^{e^{x}}, \log x, \log \log x, \cos x, \sin x),
+\]
+since, if the various arguments of~$F$ are denoted by $x$,~$y$,~$z$, $\xi$,~$\eta$, $\zeta$,~$\theta$,
+we have
+\begin{gather*}
+\frac{dy}{dx} = y,\qquad
+\frac{dz}{dx} = yz,\qquad
+\frac{d\xi}{dx} = \frac{1}{x}, \\
+%
+\frac{d\eta}{dx} = \frac{1}{x\xi},\quad
+\frac{d\zeta}{dx} = -\sqrt{1 - \zeta^{2}},\quad
+\frac{d\theta}{dx} = \sqrt{1 - \theta^{2}}.
+\end{gather*}
+The proof of the theorem does not involve ideas different in principle
+from those which have been employed continually throughout the
+preceding pages.
+
+{\small \Paragraph{6}{6.} As a final example of the manner in which these ideas may be applied,
+we shall consider the following question:
+
+`\emph{in what circumstances is
+\[
+\int R(x) \log x\, dx,
+\]
+where $R$~is rational, an elementary function?}'
+
+In the first place the integral must be of the form
+\[
+R_{0}(x, \log x)
+  + A_{1}\log R_{1}(x, \log x)
+  + A_{2}\log R_{2}(x, \log x) + \dots.
+\]
+A general consideration of the form of the differential coefficient of this
+expression, in which $\log x$ must only occur linearly and multiplied by a
+rational function, leads us to anticipate that (i)~$R_{0}(x, \log x)$ must be of the
+form
+\[
+S(x)(\log x)^{2} + T(x)\log x + U(x),
+\]
+where $S$,~$T$, and~$U$ are rational, and (ii)~$R_{1}$, $R_{2}$,~\dots\ must be rational functions
+of $x$~only; so that the integral can be expressed in the form
+\[
+S(x)(\log x)^{2} + T(x)\log x + U(x) + \sum B_{k}\log(x - a_{k}).
+\]
+%% -----File: 071.png---
+
+Differentiating, and comparing the result with the subject of integration,
+we obtain the equations
+\[
+S' = 0,\quad
+\frac{2S}{x} + T' = R,\quad
+\frac{T}{x} + U' + \sum \frac{B_{k}}{x - a_{k}} = 0.
+\]
+Hence $S$~is a constant, say~$\frac{1}{2} C$, and
+\[
+T = \int \left(R - \frac{C}{x}\right) dx.
+\]
+
+We can always determine by means of elementary operations, as in \hyperlink{4 para 4.}{\textsc{iv.},~§4},
+whether this integral is rational for any value of~$C$ or not. If not, then the
+given integral is not an elementary function. If $T$~is rational, then we must
+calculate its value, and substitute it in the integral
+\[
+U = -\int \left\{\frac{T}{x} + \sum \frac{B_{k}}{x - a_{k}}\right\} dx
+  = -\int \frac{T}{x}\, dx - \sum B_{k}\log(x - a_{k}),
+\]
+which must be rational for some value of the arbitrary constant implied in~$T$.
+We can calculate the rational part of
+\[
+\int \frac{T}{x}\, dx:
+\]
+the transcendental part must be cancelled by the logarithmic terms
+\[
+\sum B_{k}\log(x - a_{k}).
+\]
+
+The necessary and sufficient condition that the original integral should be
+an elementary function is therefore that $R$~should be of the form
+\[
+\frac{C}{x} + \frac{d}{dx} \{R_{1}(x)\},
+\]
+where $C$~is a constant and $R_{1}$~is rational. That the integral is in this case
+such a function becomes obvious if we integrate by parts, for
+\[
+\int \left(\frac{C}{x} + R_{1}'\right) \log x\, dx
+  = \tfrac{1}{2} C(\log x)^{2} + R_{1}\log x - \int \frac{R_{1}}{x}\, dx.
+\]
+
+In particular
+\[
+\Item{(i)} \int \frac{\log x}{x - a}\, dx,\qquad\qquad
+\Item{(ii)} \int \frac{\log x}{(x - a)(x - b)}\, dx,
+\]
+are not elementary functions unless in~(i) $a = 0$ and in~(ii) $b = a$. If the
+integral is elementary then the integration can always be carried out, with
+the same reservation as was necessary in the case of rational functions.
+
+It is evident that the problem considered in this paragraph is but one of
+a whole class of similar problems. The reader will find it instructive to
+formulate and consider such problems for himself.}
+%% -----File: 072.png---
+
+\Paragraph{6}{7.} It will be obvious by now that the number of classes of
+transcendental functions whose integrals are always elementary is very
+small, and that such integrals as
+\begin{alignat*}{2}
+& \int f(x, e^{x})\,dx,               && \int f(x, \log x)\,dx,\\
+& \int f(x, \cos x, \sin x)\,dx,\quad && \int f(e^{x}, \cos x, \sin x)\,dx,\\
+& \multispan{1}{\dotfill},
+\end{alignat*}
+where $f$~is algebraical, or even rational, are generally new transcendents.
+These new transcendents, like the transcendents (such as the elliptic
+integrals) which arise from the integration of algebraical functions,
+are in many cases of great interest and importance. They may often
+be expressed by means of infinite series or definite integrals, or their
+properties may be studied by means of the integral expressions which
+define them. The very fact that such a function is \emph{not} an elementary
+function in so far enhances its importance. And when such functions
+have been introduced into analysis new problems of integration arise
+in connection with them. We may enquire, for example, under what
+circumstances an elliptic integral or elliptic function, or a combination
+of such functions with elementary functions, can be integrated in finite
+terms by means of elementary and elliptic functions. But before we
+can be in a position to restate the fundamental problem of the Integral
+Calculus in any such more general form, it is essential that we should
+have disposed of the particular problem formulated in \hyperlink{3 para 1.}{Section~III}\@.
+%% -----File: 073.png---
+\BackMatter
+\Appendix{I}{BIBLIOGRAPHY}
+
+{\footnotesize The following is a list of the memoirs by Abel, Liouville and Tschebyschef
+which have reference to the subject matter of this tract.
+
+\Author{N.~H. Abel}
+
+\Bibitem[1.] `Über die Integration der Differential-Formel $\dfrac{\rho\,dx}{\sqrt{R}}$, wenn $R$~und~$\rho$ ganze
+Funktionen sind', \textit{Journal für Mathematik}, vol.~1, 1826, pp.~185--221
+(\OEuvres, vol.~1, pp.~104--144).
+
+\Bibitem[2.] `\DPtypo{Precis}{Précis} d'une théorie des fonctions elliptiques', \textit{Journal für Mathematik},
+vol.~4, 1829, pp.~236--277, 309--348 (\OEuvres, vol.~1, pp.~518--617).
+
+\Bibitem[3.] `Théorie des transcendantes elliptiques', \OEuvres, vol.~2, pp.~87--188.
+
+\Author{J. Liouville}
+
+\Bibitem[1.] `Mémoire sur la classification des transcendantes, et sur l'impossibilité
+d'exprimer les racines de certaines équations en fonction finie explicite
+des coefficients', \textit{Journal de mathématiques}, ser.~1, vol.~2, 1837,
+pp.~56--104.
+
+\Bibitem[2.] `{\Loosen Nouvelles recherches sur la détermination des intégrales dont la valeur
+est algébrique}', \ibid., vol.~3, 1838, pp.~20--24 (previously published in
+the \textit{Comptes Rendus}, 28~Aug.\ 1837).
+
+\Bibitem[3.] `Suite du mémoire sur la classification des transcendantes, et sur
+l'impossibilité d'exprimer les racines de certaines équations en fonction
+finie explicite des coefficients', \ibid., pp.~523--546\DPtypo{}{.}
+
+\Bibitem[4.] `Note sur les transcendantes elliptiques considérées comme fonctions de
+leur module', \textit{ibid.}, vol.~5, 1840, pp.~34--37.
+
+\Bibitem[5.] `Mémoire sur les transcendantes elliptiques considérées comme fonctions
+de leur module', \ibid., pp. 441--464.
+
+\Bibitem[6.] `Premier mémoire sur la détermination des intégrales dont la valeur est
+algébrique', \textit{Journal de l'École Polytechnique}, vol.~14, cahier~22, 1833,
+pp.~124--148 (also published in the \textit{Mémoires présentés par divers
+savants à l'Académie des Sciences}, vol.~5, 1838, pp.~76--151).
+
+\Bibitem[7.] `Second mémoire sur la détermination des intégrales dont la valeur est
+algébrique', \ibid., pp.~149--193 (also published as above).
+%% -----File: 074.png---
+
+\Bibitem[8.] `Mémoire sur les transcendantes elliptiques considérées comme fonctions
+de leur amplitude', \ibid., cahier~23, 1834, pp.~37--83.
+
+\Bibitem[9.] `Mémoire sur l'intégration d'une classe de fonctions transcendantes',
+\textit{Journal für Mathematik}, vol.~13, 1835, pp.~93--118.
+
+\Author{P. Tschebyschef}
+
+\Bibitem[1.] `Sur l'intégration des différentielles irrationnelles', \textit{Journal de mathématiques},
+ser.~1, vol.~18, 1853, pp.~87--111 (\OEuvres, vol.~1, pp.~147--168).
+
+\Bibitem[2.] `Sur l'intégration des différentielles qui contiennent une racine carrée
+d'une \DPtypo{polynome}{polynôme} du troisième ou du quatrième degré', \ibid., ser.~2,
+vol.~2, 1857, pp.~1--42 (\OEuvres, vol.~1, pp.~171--200; also published
+in the \textit{Mémoires de l'Académie Impériale des Sciences de St-Pétersbourg},
+ser.~6, vol.~6, 1857, pp.~203--232).
+
+\Bibitem[3.] `Sur l'intégration de la différentielle $\dfrac{x + A}{\sqrt{x^{4} + \alpha x^{3} + \beta x^{2} + \gamma x + \delta}}\,dx$', \ibid.,
+ser.~2, vol.~9, 1864, pp.~225--241 (\OEuvres, vol.~1, pp.~517--530;
+previously published in the \textit{Bulletin de l'Académie Impériale des
+Sciences de St-Pétersbourg}, vol.~3, 1861, pp.~1--12).
+
+\Bibitem[4.] `Sur l'intégration des différentielles irrationnelles', \ibid., pp.~242--246
+(\OEuvres, vol.~1, pp.~511--514; previously published in the \textit{Comptes
+Rendus}, 9~July 1860).
+
+\Bibitem[5.] `Sur l'intégration des différentielles qui contiennent une racine cubique'
+(\OEuvres, vol.~1, pp.~563--608; previously published only in Russian).
+
+Other memoirs which may be consulted are:
+
+\Author{A. Clebsch}
+
+\Bibitem[] `Über diejenigen Curven, deren Coordinaten sich als elliptische Functionen
+eines Parameters darstellen lassen', \textit{Journal für Mathematik}, vol.~64,
+1865, pp.~210--270.
+
+\Author{J. Dolbnia}
+
+\Bibitem[] `Sur les intégrales pseudo-elliptiques d'Abel', \textit{Journal de mathématiques},
+ser.~4, vol.~6, 1890, pp.~293--311.
+
+\Author{Sir A. G. Greenhill}
+
+\Bibitem[] `Pseudo-elliptic integrals and their dynamical applications', \textit{Proc.\ London
+Math.\ Soc.}, ser.~1, vol.~25, 1894, pp.~195--304.
+
+\Author{G. H. Hardy}
+
+\Bibitem[] `Properties of logarithmico-exponential functions', \textit{Proc.\ London Math.\
+Soc.}, ser.~2, vol.~10, 1910, pp.~54--90.
+
+\Author{L. Königsberger}
+
+\Bibitem[] `Bemerkungen zu Liouville's Classificirung der Transcendenten', \textit{Mathematische
+Annalen}, vol.~28, 1886, pp.~483--492.
+%% -----File: 075.png---
+
+\Author{L. Raffy}
+
+\Bibitem[] `Sur les quadratures algébriques et logarithmiques', \textit{Annales de l'École
+Normale}, ser.~3, vol.~2, 1885, pp.~185--206.
+
+\Author{K. Weierstrass}
+
+\Bibitem[] `Über die Integration algebraischer Differentiale vermittelst Logarithmen',
+\textit{Monatsberichte der Akademie der Wissenschaften zu Berlin}, 1857,
+pp.~148--157 (\textit{Werke}, vol.~1, pp.~227--232).
+
+\Author{G. Zolotareff}
+
+\Bibitem[] `Sur la méthode d'intégration de M.~Tschebyschef', \textit{Journal de mathématiques},
+ser.~2, vol.~19, 1874, pp.~161--188.
+
+Further information concerning pseudo-elliptic integrals, and degenerate
+cases of Abelian integrals generally, will be found in a number of short notes
+by Dolbnia, Kapteyn and Ptaszycki in the \textit{Bulletin des sciences mathématiques},
+and by Goursat, Gunther, Picard, Poincaré, and Raffy in the \textit{Bulletin de la
+Société Mathématique de France}, in Legendre's \textit{Traité des functions elliptiques}
+(vol.~1, ch.~26), in Halphen's \textit{Traité des fonctions elliptiques} (vol.~2, ch.~14),
+and in Enneper's \textit{Elliptische Funktionen}. The literature concerning the
+general theory of algebraical functions and their integrals is too extensive to
+be summarised here: the reader may be referred to Appell and Goursat's
+\textit{Théorie des fonctions algébriques}, and Wirtinger's article \textit{Algebraische Funktionen
+und ihre Integrale} in the \textit{Encyclopädie der Mathematischen Wissenschaften},
+\textsc{ii}~B~2.}
+%% -----File: 076.png---
+
+\Appendix{II}{ON ABEL'S PROOF OF THE THEOREM OF V., §11}
+
+{\small Abel's proof (\OEuvres, vol.~1, p.~545) is as follows\footnotemark:
+  \footnotetext{The theorem with which Abel is engaged is a very much more general
+theorem.}
+
+We have
+\[
+\psi(x, u) = 0,
+\Tag{(1)}
+\]
+where $\psi$~is an irreducible polynomial of degree~$m$ in~$u$. If we make use of the
+equation $f(x, y) = 0$, we can introduce~$y$ into this equation, and write it in the
+form
+\[
+\phi(x, y, u) = 0,
+\Tag{(2)}
+\]
+where $\phi$~is a polynomial in the three variables $x$,~$y$, and~$u$\footnotemark;
+  \footnotetext{`Or, au lieu de supposer ces coefficiens rationnels en~$x$, nous les supposerons
+  rationnels en $x$,~$y$; \emph{car cette supposition permise simplifiera beaucoup le
+  raisonnement}'.}
+and we can
+suppose~$\phi$, like~$\psi$, of degree~$m$ in~$u$ and irreducible, that is to say not
+divisible by any polynomial of the same form  which is not a constant
+multiple of~$\phi$ or itself a constant.
+
+From $f = 0$, $\phi = 0$ we deduce
+\[
+\frac{\dd f}{\dd x}
+  + \frac{\dd f}{\dd y}\, \frac{dy}{dx} = 0,\quad
+\frac{\dd\phi}{\dd x}
+  + \frac{\dd\phi}{\dd y}\, \frac{dy}{dx}
+  + \frac{\dd\phi}{\dd u}\, \frac{du}{dx} = 0;
+\]
+and, eliminating $\dfrac{dy}{dx}$, we obtain an equation of the form
+\[
+\frac{du}{dx} = \frac{\lambda(x, y, u)}{\mu(x, y, u)},
+\]
+where $\lambda$~and~$\mu$ are polynomials in $x$,~$y$, and~$u$. And in order that $u$~should
+be an integral of~$y$ it is necessary and sufficient that
+\[
+\lambda - y\mu = 0.
+\Tag{(3)}
+\]
+
+Abel now applies Lemma~(2) of \hyperlink{5 para 11.}{§11}, or rather its analogue for polynomials
+in~$u$ whose coefficients are polynomials in $x$~and~$y$, to the two polynomials $\phi$
+and $\lambda - y\mu$, and infers that \emph{all} the roots $u$,~$u'$,~\dots\ of $\phi = 0$ satisfy~\Eq{(3)}. From
+this he deduces that $u$,~$u'$,~\dots\ are all integrals of~$y$, and so that
+\[
+\frac{u + u' + \dots}{m + 1}
+\Tag{(4)}
+\]
+%% -----File: 077.png---
+is an integral of~$y$. As \Eq{(4)}~is a symmetric function of the roots of~\Eq{(2)}, it is a
+rational function of $x$~and~$y$, whence his conclusion follows\footnotemark.
+  \footnotetext{Bertrand (\textit{Calcul intégral}, ch.~5) replaces the last step in Abel's argument by
+  the observation that if $u$~and~$u'$ are both integrals of~$y$ then $u - u'$~is constant (cf.\
+  p.~39, bottom). It follows that the degree of the equation which defines~$u$ can be
+  decreased, which contradicts the hypothesis that it is irreducible.}
+
+It will be observed that the hypothesis that \Eq{(2)}~does actually involve~$y$
+is essential, if we are to avoid the absurd conclusion that $u$~is necessarily
+\emph{a rational function of $x$~only}. On the other hand it is not obvious how
+the presence of~$y$ in~$\phi$ affects the other steps in the argument.
+
+The crucial inference is that which asserts that because the equations
+$\phi = 0$ and $\lambda - y\mu = 0$, considered as equations in~$u$, have a root in common,
+and $\phi$~is irreducible, therefore $\lambda - y\mu$ is divisible by~$\phi$. This inference is
+invalid.
+
+We could only apply the lemma in this way if the equation~\Eq{(3)} were
+satisfied by one of the roots of~\Eq{(2)} \emph{identically}, that is to say for all values of
+$x$~and~$y$. But this is not the case. The equations are satisfied by the same
+value of~$u$ \emph{only when $x$~and~$y$ are connected by the equation~\Eq{(1)}}.
+
+Suppose, for example, that
+\[
+y = \frac{1}{\sqrt{1 + x}},\quad
+u = 2\sqrt{1 + x}.
+\]
+Then we may take
+\begin{align*}
+f &= (1 + x)y^{2} - 1, \\
+\psi &= u^{2} - 4(1 + x),
+\intertext{and}
+\phi &= uy - 2.
+\end{align*}
+Differentiating the equations $f = 0$ and $\phi = 0$, and eliminating~$\dfrac{dy}{dx}$, we find
+\[
+\frac{du}{dx} = \frac{u}{2(1 + x)} = \frac{\lambda}{\mu}.
+\]
+Thus
+\[
+\phi = uy - 2,\quad
+\lambda - y\mu = u - 2y(1 + x);
+\]
+and these polynomials have a common factor only in virtue of the equation
+$f = 0$.}
+%% -----File: 078.png---
+%[Blank Page]
+%% -----File: 079.png---
+%[Blank Page]
+%% -----File: 080.png---
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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% End of the Project Gutenberg EBook of The Integration of Functions of a %
+% Single Variable, by G. H. Hardy                                         %
+%                                                                         %
+% *** END OF THIS PROJECT GUTENBERG EBOOK INTEGRATION OF FUNCTIONS OF ONE VARIABLE ***
+%                                                                         %
+% ***** This file should be named 38993-t.tex or 38993-t.zip *****        %
+% This and all associated files of various formats will be found in:      %
+%         http://www.gutenberg.org/3/8/9/9/38993/                         %
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diff --git a/LICENSE.txt b/LICENSE.txt
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+This eBook, including all associated images, markup, improvements,
+metadata, and any other content or labor, has been confirmed to be
+in the PUBLIC DOMAIN IN THE UNITED STATES.
+
+Procedures for determining public domain status are described in
+the "Copyright How-To" at https://www.gutenberg.org.
+
+No investigation has been made concerning possible copyrights in
+jurisdictions other than the United States. Anyone seeking to utilize
+this eBook outside of the United States should confirm copyright
+status under the laws that apply to them.
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+Project Gutenberg (https://www.gutenberg.org) public repository for
+eBook #38993 (https://www.gutenberg.org/ebooks/38993)