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.

The standard way consists in decomposing the integrand into simple fractions:
$$\frac1{(x^2-1)^2}=\frac1{(x-1)^2(x+1)^2}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{x+1}+\frac D{(x+1)^2}.$$
Noting the integrand is an even function, we deduce that $C=-A$ and $D=B$.
Now multiply both sides by $(x^2-1)^2$: we  obtain
$$1= A(x-1)(x+1)^2+B(x+1)^2+C(x-1)^2(x+1)+D(x-1)^2,$$
and setting $x=1$ yields the relation $\;1=4B$, whence $B=D=\frac14$.
Next set $x=0$: we now get $1=-A+B+C+D=-2A+\frac12$, whence $A=\frac14$, $C=-\frac14$.
Can you proceed from here?
