I need some help finding C & D in the following integral

$\displaystyle \int\frac{2x^3-4x-8}{(x^2-x)(x^2+4)}$

$\displaystyle 2x^3-4x-8=A(x-1)(x^2+4)+B(x)(x^2+4)+[C(2x)+D](x)(x-1)$

I used x=1 to find that B=-2

I used x=0 to find that A=2

I am now trying to find C & D and am having some trouble.

When I attempt to use x=-1, I get

$\displaystyle -6=-10+-4C+2D$

How do I find C & D? Is x=-1 a poor choice? What would be a better choice?