Evaluating $\int \frac{dx}{x^2 - 2x} dx$ $$\int \frac{dx}{x^2 - 2x}$$
I know that I have to complete the square so the problem becomes.
$$\int \frac{dx}{(x - 1)^2 -1}dx$$
Then I set up my A B and C stuff
$$\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{-1}$$
With that I find $A = -1, B = -1$ and $C = 0$ which I know is wrong.
I must be setting up the $A, B, C$ thing wrong but I do not know why.

Added: "I know that I have to complete the square" is ambiguous. I interpreted it as meaning that the OP thought that completing the square was necessary to solve the problem. 
Completing the square is not a universal tool.  To find the integral efficiently, you certainly do not need to complete the square.
The simplest approach is to use partial fractions.  The bottom factors as $x(x-2)$. Find numbers $A$ and $B$ such that 
$$\frac{1}{x^2-2x}=\frac{A}{x-2}+\frac{B}{x}.$$
