I need help with integrating $\int\frac{1}{(x^2-1)^2}dx$ The problem is:
$$\int\frac{1}{(x^2-1)^2}dx$$
I tried using substitution $u=x^2-1$ but that does not bring me got results. I get an integral:
$$\frac{1}{2}\int\frac{1}{u^2\sqrt{u+1}}$$
And that does not really make things any more simple. From here I tried using partial decomposition but didn't really get anywhere.
Any help with this would be much appreciated.

$$\begin{align*}
\int \frac{1}{(x^2 - 1)^2}\,\mathrm{d}x &\equiv \int \frac{1}{\left((x + 1)\cdot(x - 1)\right)^2}\,\mathrm{d}x \\
&= \int \frac{1}{(x + 1)^2\cdot(x - 1)^2}\,\mathrm{d}x\tag{1}
\end{align*}$$
Decompose $(1)$ into partial fractions,
$$\int \frac{1}{(x - 1)^2\cdot(x + 1)^2}\,\mathrm{d}x = \frac 14\int \frac{1}{x + 1} + \frac{1}{(x + 1)^2} - \frac{1}{x - 1} + \frac{1}{(x - 1)^2}\,\mathrm{d}x\tag{2}$$
Apply linearity to split the integral and then use the substitution $u = x + 1$ for the first two terms and $u = x - 1$ for the last two terms.
