The question starts as

$\displaystyle \frac{4x}{(1+x)(1-x)^{2}}$

I've gotten this far and don't know how to solve the rest.

$\displaystyle \frac{4x}{(1+x)(1-x)^{2}}\equiv\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{(1-x)^{2}}$

$\displaystyle 4x=A(1-x)^{2}+B(1-x)(1+x)+c(1+x)$

Substitute

$\displaystyle x=1$

$\displaystyle 2=c$

Then I don't know how to solve the rest, when I substitute an x, I get a zero for both, it's completely different from this.