Partial Fraction Decomposition of quadratic factor I'm trying to break up the following equation into partial fractions:
$$\frac{1}{(x^2-1)^2}=\frac{1}{(x+1)^2(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$$
I'm ending up with two equations and four unknowns though:
$$A+C=0 \\ -A + B + C+D = 0$$
(these are the $x$ and constant coefficient variables).
How do I break this up completely?

We have
$$\frac{1}{(x^2-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$$
$$=\frac{A(x+1)(x^2-1)+B(x+1)^2+C(x-1)(x^2-1)+D(x-1)^2}{(x^2-1)^2}=$$
$$=\frac{A(x^3+x^2-x-1)+B(x^2+2x+1)+C(x^3-x^2-x+1)+D(x^2-2x+1)}{(x^2-1)^2}=$$
that is


*

*$A+C=0 \implies A=-C$

*$A+B-C+D=0\implies 2A+B+D=0$

*$-A+2B-C-2D=0\implies B=D \implies A=-B$

*$-A+B+C+D=1\implies -2A+2B=1 \implies A=-\frac14$


and then $A=-B=-C=-D=-\frac14$
