Decomposition of a rational function in partial fractions I am trying to decompose the following rational function: $\dfrac{1}{(x^2-1)^2}$ in partial fractions (in order to untegrate it later).
I have notices that $(x^2-1)^2 = (x+1)^2(x-1)^2$
Therefore $\exists A, B, C, D$ s.t:  $\dfrac{1}{(x^2-1)^2} = \dfrac{1}{(x-1)^2(x+1)^2} = \dfrac{Ax+B}{(x+1)^2}+\dfrac{Cx+D}{(x-1)^2}$
We then have $Ax+B =\left. \dfrac{1}{(x-1)^2} \right\vert _{x=-1} \implies B-A=1/4$
Same thing for Cx+D: $Cx+D =\left. \dfrac{1}{(x+1)^2} \right\vert _{x=1} \implies C+D=1/4  $
How do I find A, B, C, D from here? 

Hint: substitute $x=i $ to get two additional equations one for the real part and one for the imaginary part
