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.

$$\begin{align} & {} \quad \int \frac{dx}{x^2 - 2x}\\ &=\int \frac{dx}{(x - 1)^2 -1}dx\\ &=\int \frac{dx}{(x - 1-1) (x-1+1)} \\ &=\int \frac{dx}{x(x-2)}\end{align}$$
The rest is easy (partial fractions):
$$\frac 12 \left[\int \frac {dx}{x-2} - \int \frac {dx}x\right]=\frac 12 \ln|x-2| -\frac 12 \ln |x| +C$$
