Partial Fraction Decomposition of $\frac{x^2}{(x-1)(x+1)^2}$ I am suppose to find the Partial Fraction Decomposition of
$$\dfrac{x^2}{(x-1)(x+1)^2}$$
My work so far:
$$\dfrac{x^2}{(x-1)(x+1)^2} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}$$
Thus,
$$ x^2 = A(x+1)^3 + B(x-1)(x+1)^2 + C(x+1)(x-1)$$
With x = 0 we get,
$$ 0 = A - B - C $$
I am not sure what to do after this, as plugging in the remaining factors (x = -1, 1) Does not help me find the unknown coefficients.
Any help would be appreciated. Thanks
Edit. With $x = 1$, 
$$1 = A + 0B + 0C$$
$$A = 1$$ 
With $x = -1$
$$1 = 0A + 0B + 0C$$
$$1 = 0$$

After your "thus" it must be
$$x^2=A(x+1)^2+B(x-1)(x+1)+C(x-1)$$
and not what you wrote there. Now substitute $\;x=1\;$ in both sides and get
$$1^2=A\cdot2^2\implies A=\frac14$$
and then substitute $\;x=-1\;$ and get ...etc.
