∫ dx/(2x-1)³

Here's what I have so far:

1 = A(2x-1)² + B(2x-1) + C

[solve for c]

x=1/2

1= A(0)² + B(0) + C

1=C

But in order to solve for A or B, I'd have to use x=1/2, but that makes BOTH terms 0.

Am I supposed to plug in a random # for x and then get a system of equations, or is there an easier way I'm missing?