$\displaystyle integrate: \frac{x^3 + 3x^2 + 5x -9}{(x-5)(x+1)^3} dx$

I worked it down to $\displaystyle x^3 + 3x^2 + 5x -9 = A(x+1)^3 + B(x-5)(x+1)^2 + c(x-5)(x+1) + D(x-5)$

and then

x = 5

a= 1

x = -1

-12 = b

whats confusing me is that if i let x = -1 for c i would get -12 = C0 which is -12 = 0

is my equation wrong?