132-139. Integration of algebraical functions. Integration by rationalisation. Integration by parts

132. Algebraical Functions.

We naturally pass on next to the question of the integration of algebraical functions. We have to consider the problem of integrating $y$, where $y$ is an algebraical function of $x$. It is however convenient to consider an apparently more general integral, viz. $$\int R(x,y)dx,$$ where $R(x,y)$ is any rational function of $x$ and $y$. The greater generality of this form is only apparent, since (Ex. XIV. 6) the function $R(x,y)$ is itself an algebraical function of $x$. The choice of this form is in fact dictated simply by motives of convenience: such a function as $$\frac{px+q+\sqrt{a{x}^2+2bx+c}}{px+q–\sqrt{a{x}^2+2bx+c}}$$ is far more conveniently regarded as a rational function of $x$ and the simple algebraical function $\sqrt{a{x}^2+2bx+c}$, than directly as itself an algebraical function of $x$.

133. Integration by substitution and rationalisation.

It follows from equation (3) of § 130 that if $\int \psi (x)dx=\varphi (x)$ then $$\begin{array}{}\text{(1)}& \int \psi \{f(t)\}{f}^{\prime }(t)dt=\varphi \{f(t)\}.\end{array}$$

This equation supplies us with a method for determining the integral of $\psi (x)$ in a large number of cases in which the form of the integral is not directly obvious. It may be stated as a rule as follows: put $x=f(t)$, where $f(t)$ is any function of a new variable $t$ which it may be convenient to choose; multiply by $f^{\prime }(t)$, and determine $ifpossible$ the integral of $\psi \{f(t)\}{f}^{\prime }(t)$; express the result in terms of $x$. It will often be found that the function of $t$ to which we are led by the application of this rule is one whose integral can easily be calculated. This is always so, for example, if it is a rational function, and it is very often possible to choose the relation between $x$ and $t$ so that this shall be the case. Thus the integral of $R(\sqrt{x})$, where $R$ denotes a rational function, is reduced by the substitution $x=t^2$ to the integral of $2tR(t^2)$, to the integral of a rational function of $t$. This method of integration is called integration by rationalisation, and is of extremely wide application.

Its application to the problem immediately under consideration is obvious. If we can find a variable $t$ such that $x$ and $y$ are both rational functions of $t$, say $x=R_1(t)$, $y=R_2(t)$, then $$\int R(x,y)dx=\int R\{R_1(t),R_2(t)\}{R}_1^{\prime }(t)dt,$$ and the latter integral, being that of a rational function of $t$, can be calculated by the methods of § 130.

It would carry us beyond our present range to enter upon any general discussion as to when it is and when it is not possible to find an auxiliary variable $t$ connected with $x$ and $y$ in the manner indicated above. We shall consider only a few simple and interesting special cases.

134. Integrals connected with conics.

Let us suppose that $x$ and $y$ are connected by an equation of the form $$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0;$$ in other words that the graph of $y$, considered as a function of $x$ is a conic. Suppose that $(\xi ,\eta )$ is any point on the conic, and let $x–\xi =X$, $y–\eta =Y$. If the relation between $x$ and $y$ is expressed in terms of $X$ and $Y$, it assumes the form $$a{X}^2+2hXY+b{Y}^2+2GX+2FY=0,$$ where $F=h\xi +b\eta +f$, $G=a\xi +h\eta +g$. In this equation put $Y=tX$. It will then be found that $X$ and $Y$ can both be expressed as rational functions of $t$, and therefore $x$ and $y$ can be so expressed, the actual formulae being $$x–\xi =-\frac{2(G+Ft)}{a+2ht+b{t}^2},y–\eta =-\frac{2t(G+Ft)}{a+2ht+b{t}^2}.$$ Hence the process of rationalisation described in the last section can be carried out.

The reader should verify that $$hx+by+f=-\frac{1}{2}(a+2ht+b{t}^2)\frac{dx}{dt},$$ so that $$\int \frac{dx}{hx+by+f}=-2\int \frac{dt}{a+2ht+b{t}^2}.$$

When $h^2>ab$ it is in some ways advantageous to proceed as follows. The conic is a hyperbola whose asymptotes are parallel to the lines $$a{x}^2+2hxy+b{y}^2=0,$$ or $$b(y–\mu x)(y–{\mu }^{\prime }x)=0,$$ say. If we put $y–\mu x=t$, we obtain $$y–\mu x=t,y–{\mu }^{\prime }x=-\frac{2gx+2fy+c}{bt},$$ and it is clear that $x$ and $y$ can be calculated from these equations as rational functions of $t$. We shall illustrate this process by an application to an important special case.

135. The integral $\int \frac{dx}{\sqrt{a{x}^2+2bx+c}}$.

Suppose in particular that $y^2=a{x}^2+2bx+c$, where $a>0$. It will be found that, if we put $y+x\sqrt{a}=t$, we obtain $$2\frac{dx}{dt}=\frac{(t^2+c)\sqrt{a}+2bt}{(t\sqrt{a}+b{)}^2},2y=\frac{(t^2+c)\sqrt{a}+2bt}{t\sqrt{a}+b},$$ and so $$\begin{array}{}\text{(1)}& \int \frac{dx}{y}=\int \frac{dt}{t\sqrt{a}+b}=\frac{1}{\sqrt{a}}\log |x\sqrt{a}+y+\frac{b}{\sqrt{a}}|.\end{array}$$ If in particular $a=1$, $b=0$, $c=a^2$, or $a=1$, $b=0$, $c=-a^2$, we obtain $$\begin{array}{}\text{(2)}& \int \frac{dx}{\sqrt{x^2+a^2}}=\log \{x+\sqrt{x^2+a^2}\},\int \frac{dx}{\sqrt{x^2–a^2}}=\log |x+\sqrt{x^2–a^2}|,\end{array}$$ equations whose truth may be verified immediately by differentiation. With these formulae should be associated the third formula $$\begin{array}{}\text{(3)}& \int \frac{dx}{\sqrt{a^2–x^2}}=\arcsin (x/a),\end{array}$$ which corresponds to a case of the general integral of this section in which $a<0$. In it is supposed that $a>0$; if $a<0$ then the integral is $\arcsin (x/|a|)$ (cf. § 119). In practice we should evaluate the general integral by reducing it (as in the next section) to one or other of these standard forms.

The formula (3) appears very different from the formulae (2): the reader will hardly be in a position to appreciate the connection between them until he has read Ch. X.

136. The integral $\int \frac{\lambda x+\mu }{\sqrt{a{x}^2+2bx+c}}dx$.

This integral can be integrated in all cases by means of the results of the preceding sections. It is most convenient to proceed as follows. Since $$\begin{array}{c}\lambda x+\mu =(\lambda /a)(ax+b)+\mu –(\lambda b/a),\\ \int \frac{ax+b}{\sqrt{a{x}^2+2bx+c}}dx=\sqrt{a{x}^2+2bx+c},\end{array}$$ we have $$\int \frac{(\lambda x+\mu )dx}{\sqrt{a{x}^2+2bx+c}}=\frac{\lambda }{a}\sqrt{a{x}^2+2bx+c}+\left(\mu –\frac{\lambda b}{a}\right)\int \frac{dx}{\sqrt{a{x}^2+2bx+c}}.$$

In the last integral $a$ may be positive or negative. If $a$ is positive we put $x\sqrt{a}+(b/\sqrt{a})=t$, when we obtain $$\frac{1}{\sqrt{a}}\int \frac{dt}{\sqrt{t^2+\kappa }},$$ where $\kappa =(ac–b^2)/a$. If $a$ is negative we write $A$ for $-a$ and put $x\sqrt{A}–(b/\sqrt{A})=t$, when we obtain $$\frac{1}{\sqrt{-a}}\int \frac{dt}{\sqrt{-\kappa –t^2}}.$$

It thus appears that in any case the calculation of the integral may be made to depend on that of the integral considered in § 135, and that this integral may be reduced to one or other of the three forms $$\int \frac{dt}{\sqrt{t^2+a^2}},\int \frac{dt}{\sqrt{t^2–a^2}},\int \frac{dt}{\sqrt{a^2–t^2}}.$$

137. The integral $\int (\lambda x+\mu )\sqrt{a{x}^2+2bx+c}dx$.

In exactly the same way we find $$\begin{array}{c}\int (\lambda x+\mu )\sqrt{a{x}^2+2bx+c}dx\\ =\left(\frac{\lambda }{3a}\right)(a{x}^2+2bx+c{)}^{3/2}+\left(\mu –\frac{\lambda b}{a}\right)\int \sqrt{a{x}^2+2bx+c}dx;\end{array}$$ and the last integral may be reduced to one or other of the three forms $$\int \sqrt{t^2+a^2}dt,\int \sqrt{t^2–a^2}dt,\int \sqrt{a^2–t^2}dt.$$ In order to obtain these integrals it is convenient to introduce at this point another general theorem in integration.

138. Integration by parts.

The theorem of integration by parts is merely another way of stating the rule for the differentiation of a product proved in § 113. It follows at once from Theorem (3) of § 113 that $$\int f^{\prime }(x)F(x)dx=f(x)F(x)–\int f(x)F^{\prime }(x)dx.$$ It may happen that the function which we wish to integrate is expressible in the form $f^{\prime }(x)F(x)$, and that $f(x)F^{\prime }(x)$ can be integrated. Suppose, for example, that $\varphi (x)=x\psi (x)$, where $\psi (x)$ is the second derivative of a known function $\chi (x)$. Then $$\int \varphi (x)dx=\int x\chi ”(x)dx=x{\chi }^{\prime }(x)–\int {\chi }^{\prime }(x)dx=x{\chi }^{\prime }(x)–\chi (x).$$

We can illustrate the working of this method of integration by applying it to the integrals of the last section. Taking $$f(x)=ax+b,F(x)=\sqrt{a{x}^2+2bx+c}=y,$$ we obtain $$\begin{array}{rl}a\int ydx& =(ax+b)y–\int \frac{(ax+b{)}^2}{y}dx\\ & =(ax+b)y–a\int ydx+(ac–b^2)\int \frac{dx}{y},\end{array}$$ so that $$\int ydx=\frac{(ax+b)y}{2a}+\frac{ac–b^2}{2a}\int \frac{dx}{y};$$ and we have seen already (§ 135) how to determine the last integral.

139. The general integral $\int R(x,y)dx$, where $y^2=a{x}^2+2bx+c$.

The most general integral, of the type considered in § 134, and associated with the special conic $y^2=a{x}^2+2bx+c$, is $$\begin{array}{}\text{(1)}& \int R(x,\sqrt{X})dx,\end{array}$$ where $X=y^2=a{x}^2+2bx+c$. We suppose that $R$ is a real function.

The subject of integration is of the form $P/Q$, where $P$ and $Q$ are polynomials in $x$ and $\sqrt{X}$. It may therefore be reduced to the form $$\frac{A+B\sqrt{X}}{C+D\sqrt{X}}=\frac{(A+B\sqrt{X})(C–D\sqrt{X})}{C^2–D^{2}X}=E+F\sqrt{X},$$ where $A$, $B$, … are rational functions of $x$. The only new problem which arises is that of the integration of a function of the form $F\sqrt{X}$, or, what is the same thing, $G/\sqrt{X}$, where $G$ is a rational function of $x$. And the integral $$\begin{array}{}\text{(2)}& \int \frac{G}{\sqrt{X}}dx\end{array}$$ can always be evaluated by splitting up $G$ into partial fractions. When we do this, integrals of three different types may arise.

(i) In the first place there may be integrals of the type $$\begin{array}{}\text{(3)}& \int \frac{x^m}{\sqrt{X}}dx,\end{array}$$ where $m$ is a positive integer. The cases in which $m=0$ or $m=1$ have been disposed of in § 136. In order to calculate the integrals corresponding to larger values of $m$ we observe that $$\frac{d}{dx}(x^{m-1}\sqrt{X})=(m–1)x^{m-2}\sqrt{X}+\frac{(ax+b)x^{m-1}}{\sqrt{X}}=\frac{\alpha x^m+\beta x^{m-1}+\gamma x^{m-2}}{\sqrt{X}},$$ where $\alpha$, $\beta$, $\gamma$ are constants whose values may be easily calculated. It is clear that, when we integrate this equation, we obtain a relation between three successive integrals of the type (3). As we know the values of the integral for $m=0$ and $m=1$, we can calculate in turn its values for all other values of $m$.

(ii) In the second place there may be integrals of the type $$\begin{array}{}\text{(4)}& \int \frac{dx}{(x–p{)}^m\sqrt{X}},\end{array}$$ where $p$ is real. If we make the substitution $x–p=1/t$ then this integral is reduced to an integral in $t$ of the type (3).

(iii) Finally, there may be integrals corresponding to complex roots of the denominator of $G$. We shall confine ourselves to the simplest case, that in which all such roots are simple roots. In this case (cf. § 130) a pair of conjugate complex roots of $G$ gives rise to an integral of the type $$\begin{array}{}\text{(5)}& \int \frac{Lx+M}{(A{x}^2+2Bx+C)\sqrt{a{x}^2+2bx+c}}dx.\end{array}$$

In order to evaluate this integral we put $$x=\frac{\mu t+\nu }{t+1},$$ where $\mu$ and $\nu$ are so chosen that $$a\mu \nu +b(\mu +\nu )+c=0,A\mu \nu +B(\mu +\nu )+C=0;$$ so that $\mu$ and $\nu$ are the roots of the equation $$(aB–bA){\xi }^2–(cA–aC)\xi +(bC–cB)=0.$$ This equation has certainly real roots, for it is the same equation as equation (1) of Ex. XLVI. 12; and it is therefore certainly possible to find real values of $\mu$ and $\nu$ fulfilling our requirements.

It will be found, on carrying out the substitution, that the integral (5) assumes the form $$\begin{array}{}\text{(6)}& H\int \frac{tdt}{(\alpha t^2+\beta )\sqrt{\gamma t^2+\delta }}+K\int \frac{dt}{(\alpha t^2+\beta )\sqrt{\gamma t^2+\delta }}.\end{array}$$ The second of these integrals is rationalised by the substitution $$\frac{t}{\sqrt{\gamma t^2+\delta }}=u,$$ which gives $$\int \frac{dt}{(\alpha t^2+\beta )\sqrt{\gamma t^2+\delta }}=\int \frac{du}{\beta +(\alpha \delta –\beta \gamma )u^2}.$$ Finally, if we put $t=1/u$ in the first of the integrals (6), it is transformed into an integral of the second type, and may therefore be calculated in the manner just explained, viz. by putting $u/\sqrt{\gamma +\delta u^2}=u$, i.e. $1/\sqrt{\gamma t^2+\delta }=v$.¹