Evaluate the following integral:

Let, $I=\int \frac{1}{\left(x^{4}-1\right)} d x$

Let $\frac{1}{\left(x^{4}-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x^{2}+1}$b

$\Rightarrow 1=A(x-1)\left(x^{2}+1\right)+B(x+1)\left(x^{2}+1\right)+c(x+1)(x-1)$

For, $\mathrm{x}=1, \mathrm{~B}=\frac{1}{4}$

For, $x=-1, A=\frac{1}{4}$

For, $x=0, A=-\frac{1}{2}$

$\therefore \mathrm{I}=-\frac{1}{4} \int \frac{\mathrm{dx}}{\mathrm{x}+1}+\frac{1}{4} \int \frac{\mathrm{dx}}{\mathrm{x}-1}-\frac{1}{2} \int \frac{\mathrm{dx}}{\mathrm{x}^{2}+1}$

$\Rightarrow I=-\frac{1}{4} \ln |(x+1)|+\frac{1}{4} \ln |x-1|-\frac{1}{2} \tan ^{-1} x+$

$\Rightarrow I=\frac{1}{4} \ln \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+c$

So, $\int \frac{1}{\left(x^{4}-1\right)} d x=\frac{1}{4} \ln \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+c$