Partial fraction $\int \frac{x^2 + 11x dx}{(x-1)(x+1)^2}$ I have been using the cover up method from this 
video lecture
$$\int \frac{x^2 + 11x}{(x-1)(x+1)^2} dx$$
$$\frac{x^2 + 11x }{(x-1)(x+1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{x-1}$$
What do I do with the double $x-1$ denominator? With the cover up method it seems useless. I can get that $A$ is $3$, $B$ might be 5 but then so is $C$? How do I use the cover up method to solve this?

You need to decompose as follows
$${x^2 + 11 x\over (x-1)(x+1)^2}  = {A\over x + 1} + {B\over x - 1} + {C\over (x+1)^2}.$$
