Tedious fraction decomposition integral $\int\frac{1}{(x^2-1)^2} \, dx$

Question

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?

Answer 1

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. $$\begin{align}\frac{1}{(x^2-1)^2} &= \left[\frac{1}{(x-1)(x+1)}\right]^2 = \frac14\left[\frac{1}{x-1} - \frac{1}{x+1}\right]^2\\ &= \frac14\left[\frac{1}{(x-1)^2} - \frac{2}{(x-1)(x+1)} + \frac{1}{(x+1)}^2\right]\\ &= \frac14\left[\frac{1}{(x-1)^2} - \left(\frac{1}{x-1} - \frac{1}{x+1}\right) + \frac{1}{(x+1)^2}\right]\end{align}$$

Answer 2

Alternatively start from $\frac{1}{(x^2-a)^2} = \frac{\partial}{\partial a} \frac{1}{(x^2-a)}$, then determine $\int\frac{1}{(x^2-a)} \, dx$ (this is straight forward to put the integrand into partial fractions), differentiate this wrt $a$ then set $a=1$.