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?
 A: In general, for any factor in the denominator $(ax + b)^n$, where $n > 1$, we need to account for each multiplicity of factors: $$\frac{A}{ax + b} + \frac{B}{(ax + b)^2} + \cdots + \frac Z{(ax + b)^n}$$
So...You need to decompose as follows:
$${x^2 + 11 x\over (x-1){\bf (x+1)^2}}  = {A\over x - 1} + {B\over {\bf x + 1}} + {C\over {\bf (x+1)^2}}.$$
Then essentially we solve for $$A(x+1)^2 + B(x-1)(x+ 1) + C(x-1) = x^2 + 11x + 0$$ by matching up coefficients. (It's like how we find the greatest common denominator, which is then set equal to the numerator of the original integrand.
Simpler yet: 
$x = 1 \implies 4A = 12 \implies A = 3.\;$ 
Putting $x = -1 \implies -2C = -10 \implies C = 5$. 
Now just solve for B: $x = 0 \implies A - B - C = 0 \implies B = A-C = 3-5 = -2$.
This gives us $${x^2 + 11 x\over (x-1)(x+1)^2}  = {3\over x - 1} - {2\over x + 1} + {5\over (x+1)^2}$$
A: The third denominator should be $(x+1)^2$.  When you have repeated factors in the denominator, you need one term with each power.  The numerator can still be a constant.
A: You need to decompose as follows
$${x^2 + 11 x\over (x-1)(x+1)^2}  = {A\over x + 1} + {B\over x - 1} + {C\over (x+1)^2}.$$
