<s>[INST] 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.
 [/INST] I suppose that you could have also used trigonometric substitutions. Completing the square as you have done, we get that 
$$\displaystyle \int \frac{dx}{x^2-2x} = \int \frac{dx}{(x-1)^2 - 1}.$$
We now let $x-1 = \sec \theta.$ Notice then that $dx = \tan \theta \sec \theta  \ d\theta.$ We now have 
$$ \int \frac{dx}{(x-1)^2 - 1} = \int \frac{\tan \theta \sec \theta \ d\theta}{\sec^2\theta-1} = \int \frac{\tan \theta \sec \theta \ d\theta}{\tan^2 \theta} = \int \frac{\sec \theta \ d\theta}{\tan \theta} = \int \csc \theta \ d\theta. $$
And,
$$ \int \csc \theta \ d\theta = \ln (\csc \theta - \cot \theta) + C = \ln \left(\frac{\sqrt{x-2}}{\sqrt{x}} \right) + C,$$ as desired. 
(This is because from $x-1 = \sec \theta,$  we get $1/(x-1) = \cos \theta$ and thus $\sin \theta = (\sqrt{x^2-2x})/(x-1)$, and we just make the necessary substitutions in $\csc \theta - \cot \theta$).
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