Evaluate the integral:

Solution:

$I=\int \frac{a x^{2}+b x}{x^{4}+c^{2}} d x$

As we can see that there is a term of $x^{3}$ in numerator and derivative of $x^{4}$ is also $4 x^{3} .$ So there is a chance that we can make substitution for $x^{4}+c^{2}$ and I can be reduced to a fundamental integration but there is also a $x$ term present. So it is better to break this integration.

$I=\int \frac{a x^{3}}{x^{4}+c^{2}} d x+\int \frac{b x}{x^{4}+c^{2}} d x=I_{1}+I_{2} \ldots$ eqn 1

$I_{1}=\int \frac{a x^{2}}{x^{4}+c^{2}} d x=\frac{a}{4} \int \frac{4 x^{2}}{x^{4}+c^{2}} d x$'

As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{4}+\mathrm{c}^{2}\right)=4 \mathrm{x}^{3}$

To make the substitution, $I_{1}$ can be rewritten as

$I_{1}=\frac{a}{4} \int \frac{4 x^{3}}{x^{4}+c^{2}} d x$

$\therefore$ Let, $x^{4}+c^{2}=u$

$\Rightarrow \mathrm{du}=4 \mathrm{x}^{3} \mathrm{dx}$

$I_{1}$ is reduced to simple integration after substituting $u$ and du as:

$I_{1}=\frac{a}{4} \int \frac{d u}{u}=\frac{a}{4} \log |u|+C$

$\therefore I_{1}=\frac{a}{4} \log \left|x^{4}+c^{2}\right|+C \ldots e q n 2$

As,

$I_{2}=\int \frac{b x}{x^{4}+c^{2}} d x$

$\because$ we have derivative of $x^{2}$ in numerator and term of $x^{2}$ in denominator. So we can apply method of substitution here also.

As, $I_{2}=\int \frac{b x}{\left(x^{2}\right)^{2}+c^{2}} d x$

Let, $x^{2}=v$

$\Rightarrow \mathrm{dv}=2 \mathrm{x} \mathrm{dx}$

$\because I_{2}=\frac{\mathrm{b}}{2} \int \frac{2 \mathrm{x}}{\left(\mathrm{x}^{2}\right)^{2}+\mathrm{c}^{2}} \mathrm{dx}=\frac{\mathrm{b}}{2} \int \frac{\mathrm{dv}}{(\mathrm{v})^{2}+\mathrm{c}^{2}}$

As denominator doesn't have any square root term. So one of the following two integrals will solve the problem.

i) $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$ ii) $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$

$\mathrm{I}_{2}$ matches with $\int \frac{1}{\mathrm{x}^{2}+\mathrm{a}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$

$\therefore \mathrm{I}_{2}=\frac{\mathrm{b}}{2} \frac{1}{\mathrm{c}} \tan ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)+\mathrm{K}=\frac{\mathrm{b}}{2 \mathrm{c}} \tan ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)+\mathrm{K}$

$\Rightarrow I_{2}=\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+K \ldots e q n 3$

From eqn 1, we have:

$I=I_{1}+I_{2}$

Using eqn 2 and 3 , we get -

$I=\frac{a}{4} \log \left|x^{4}+c^{2}\right|+\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+K \ldots \ldots a n s$