Question

Evaluate: $\int \frac{x}{x^4-1}dx$

Answer 1

$\int \frac{x}{x^4-1}dx$
$=\frac{1}{2}\int \frac{2x}{{\left(x^2\right)}^2-1}dx$

Substituting $x^2=t\Rightarrow 2xdx=dt$

$=\frac{1}{2}\int \frac{2x}{{\left(x^2\right)}^2-1}dx=\frac{1}{2}\int \frac{1}{t^2-1}dt$
$=\frac{1}{2}\left[\frac{1}{2}\log \left|\frac{t-1}{t+1}\right|\right]+c$
$[\because \int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+c]$
$=\frac{1}{4}\log \left|\frac{x^2-1}{x^2+1}\right|+c$ $[\because x^2=t]$

Where, $c$ is constant of integration.
Hence, the required integration is $\frac{1}{4}\log \left|\frac{x^2-1}{x^2+1}\right|+c$