Let $\frac{1}{x^4-1}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x-1}+\frac{\mathrm{C} x+\mathrm{D}}{x^2+1}$
$\Rightarrow 1 \equiv \mathrm{A}(x-1)\left(x^2+1\right)+\mathrm{B}(x+1)\left(x^2+1\right)$
$+(\mathrm{C} x+\mathrm{D})(x+1)(x-1) \quad \ldots(i)$
Put $x=-1,1$ in (i), we get : $\mathrm{A}=\frac{-1}{4} \& \mathrm{~B}=\frac{1}{4}$
Comparing coefficients $\mathrm{C}=0$ and $\mathrm{D}=-\frac{1}{2}$
$\mathrm{I}=-\frac{1}{4} \int \frac{d x}{x+1}+\frac{1}{4} \int \frac{1}{x-1} d x-\frac{1}{2} \int \frac{1}{\left(x^2+1\right)} d x$
$=\frac{1}{4} \log \mid \frac{x-1}{x+1}-\frac{1}{2} \tan ^{-1} x+\mathrm{C}$