Evaluate the following integral:

Solution:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

$\frac{\mathrm{x}}{\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}-\mathrm{b}^{2}\right)}=\frac{\mathrm{Ax}+\mathrm{B}}{\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\left(\mathrm{x}^{2}-\mathrm{b}^{2}\right)} \ldots \ldots$ (i)

$\Rightarrow \frac{\mathrm{x}}{\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}-\mathrm{b}^{2}\right)}=\frac{(\mathrm{Ax}+\mathrm{B})\left(\mathrm{x}^{2}-\mathrm{b}^{2}\right)+(\mathrm{Cx}+\mathrm{D})\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)}{\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}-\mathrm{b}^{2}\right)}$

$\Rightarrow x=(A x+B)\left(x^{2}-b^{2}\right)+(C x+D)\left(x^{2}-a^{2}\right)$

$\Rightarrow x=A x^{3}-A b^{2} x+B x^{2}-b^{2} B+C x^{3}-a^{2} C x+D x^{2}-a^{2} D$

$\Rightarrow \mathrm{x}=(\mathrm{A}+\mathrm{C}) \mathrm{x}^{3}+(\mathrm{B}+\mathrm{D}) \mathrm{x}^{2}+\left(-\mathrm{Ab}^{2}-\mathrm{Ca}^{2}\right) \mathrm{x}+\left(-\mathrm{b}^{2} \mathrm{~B}\right.$

$\left.-\mathrm{a}^{2} \mathrm{D}\right) \ldots \ldots$ (ii)

By equating similar terms, we get

$A+C=0 \Rightarrow A=-C \ldots \ldots \ldots \ldots$ (iii)

$B+D=0 \Rightarrow B=-D \ldots \ldots \ldots \ldots$ (iv)

$-\mathrm{Ab}^{2}-\mathrm{Ca}^{2}=1$

$\Rightarrow-(-C) b^{2}-C a^{2}=1($ from equation(iii))

$\Rightarrow \quad C=\frac{1}{b^{2}-a^{2}} \ldots \ldots \ldots \ldots \ldots(v)$

$-b^{2} B-a^{2} D=0$

$\Rightarrow-b^{2}(-D)-a^{2} D=0$

$\Rightarrow D=0$

So equation(iv) becomes $B=0$

So equation (iii) becomes, $A=-\frac{1}{b^{2}-a^{2}}$

We put the values of $A, B, C$, and $D$ values back into our partial fractions in equation (i) and replace this as the integrand. We get

$\int \frac{x}{\left(x^{2}-a^{2}\right)\left(x^{2}-b^{2}\right)} d x$

$\Rightarrow \int\left[\frac{A x+B}{\left(x^{2}-a^{2}\right)}+\frac{C x+D}{\left(x^{2}-b^{2}\right)}\right] d x$

$\Rightarrow \int\left[\frac{\left(-\frac{1}{b^{2}-a^{2}}\right) x+0}{\left(x^{2}-a^{2}\right)}+\frac{\left(\frac{1}{b^{2}-a^{2}}\right) x+0}{\left(x^{2}-b^{2}\right)}\right] d x$

Split up the integral,

$\Rightarrow-\frac{1}{b^{2}-a^{2}} \int \frac{1}{\left(x^{2}-a^{2}\right)} d x+\frac{1}{b^{2}-a^{2}} \int \frac{1}{\left(x^{2}-b^{2}\right)} d x$

Let substitute

$u=x^{2}-a^{2} \Rightarrow d u=2 d x$

$v=x^{2}-b^{2} \Rightarrow d v=2 d x$, so the above equation becomes,

$\Rightarrow-\frac{1}{b^{2}-a^{2}} \int \frac{\frac{1}{u} d u}{2}+\frac{1}{b^{2}-a^{2}} \int \frac{\frac{1}{v} d v}{2}$

$\Rightarrow-\frac{1}{2\left(b^{2}-a^{2}\right)} \int \frac{1}{u} d u+\frac{1}{2\left(b^{2}-a^{2}\right)} \int \frac{1}{v} d v$

On integrating we get

$\Rightarrow-\frac{1}{2\left(b^{2}-a^{2}\right)} \log |u|+\frac{1}{2\left(b^{2}-a^{2}\right)} \log |v|+C$

Substituting back, we get

$\Rightarrow \frac{1}{2\left(b^{2}-a^{2}\right)}\left[\log \left|x^{2}-b^{2}\right|-\log \left|x^{2}-a^{2}\right|\right]+C$

Applying the logarithm rule we get

$\Rightarrow \frac{1}{2\left(b^{2}-a^{2}\right)}\left[\log \left|\frac{x^{2}-b^{2}}{x^{2}-a^{2}}\right|\right]+C$

Note: the absolute value signs account for the domain of the natural log function $(x>0)$.

Hence,

$\int \frac{x}{\left(x^{2}-a^{2}\right)\left(x^{2}-b^{2}\right)} d x=\frac{1}{2\left(b^{2}-a^{2}\right)}\left[\log \left|\frac{x^{2}-b^{2}}{x^{2}-a^{2}}\right|\right]+C$