Tedious fraction decomposition integral $\int\frac{1}{(x^2-1)^2} \, dx$ I'm having this integral to resolve using fraction decomposition:
$$\int\frac{1}{(x^2-1)^2} \, dx$$
This gives me the following: 
$$\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} = \frac{1}{(x^2-1)^2}$$ 
Which results in a 4 variables system of equation...
Is there a quick way to resolve this that I'm missing out?

You can work out the partial fraction decomposition of the integrand by repeat application of the identify:
$$\frac{1}{(x-a)(x-b)} = \frac{1}{(b-a)}\left[\frac{1}{x-a} - \frac{1}{x-b}\right]$$
This is especially useful when $a,b$ are small integers.
$$\begin{align}\frac{1}{(x^2-1)^2}
&= \left[\frac{1}{(x-1)(x+1)}\right]^2 =
\frac14\left[\frac{1}{x-1} - \frac{1}{x+1}\right]^2\\
&= \frac14\left[\frac{1}{(x-1)^2} - \frac{2}{(x-1)(x+1)} + \frac{1}{(x+1)}^2\right]\\
&= \frac14\left[\frac{1}{(x-1)^2} - \left(\frac{1}{x-1} - \frac{1}{x+1}\right) + \frac{1}{(x+1)^2}\right]\end{align}$$
