Question: Evaluate the integral $$ \int \frac{dx}{(x^4 - 1)^2} $$
I solved this problem using partial fractions, to obtain the answer
$$ \frac{1}{16}\left( \frac{4x}{x^4 - 1} + 6 \tan^{-1}(x) + 3\log\left(\frac{1+x}{1-x}\right) \right)$$
However the solution was super-long, and so I hoped of finding other methods of solving it by shorter methods. So far I have tried substituting $\sqrt{\sec t}$ for $x$ and it didn't seem to work.
I have also tried substituting $x$ for $\frac{1}{t}$ so that $dx$ = $-\frac{1}{t^2}$, which yields,
$$ \int \frac{-t^8 dt}{t^2 (t^4 - 1)^2} = \int \frac{-t^6 dt}{(t^4 - 1)^2} $$
and I couldn't figure how could I proceed further from this, if it is a plausible approach.
I would like some hints as for how could I proceed through this problem to obtain a shorter solution. Thanks.