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.


My book is telling me that I have to complete the square

$$I=\begin{eqnarray*}
\int \frac{dx}{x^{2}-2x} &=&\int \frac{dx}{\left( x-1\right) ^{2}-1}\overset{
u=x-1}{=}\int \frac{1}{u^{2}-1}\,du=-\text{arctanh }u+C
\end{eqnarray*},$$
$$\tag{1}$$
where I have used the substitution $u=x-1$ and the standard derivative $$\frac{d}{du}\text {arctanh}=\frac{1}{1-u^{2}}\tag{2}$$
You just need to substitute $u=x-1$ to write $\text{arctanh }u$ in terms of $x$.
Added 2: Remark. If we use the logarithmic representation of the inverse hyperbolic function $\text{arctanh }u$
$$\begin{equation*}
\text{arctanh }u=\frac{1}{2}\ln \left( u+1\right) -\frac{1}{2}\ln \left(
1-u\right),\qquad (\text{real for }|u|<1)\tag{3} 
\end{equation*}$$
we get for $u=x−1 $
$$\begin{eqnarray*}
I &=&-\text{arctanh }u+C=-\text{arctanh }\left( x-1\right) +C \\
&=&-\frac{1}{2}\ln x+\frac{1}{2}\ln \left( 2-x\right) +C \\
&=&\frac{1}{2}\left( \ln \frac{2-x}{x}\right) +C\qquad (0<x<2).
\end{eqnarray*}\tag{4}$$
Added. If your book does require using partial fractions then you can proceed as follows
$$\begin{equation*}
\int \frac{1}{u^{2}-1}\,du=\int \frac{1}{\left( u-1\right) \left( u+1\right) 
}\,du=\int \frac{1}{2\left( u-1\right) }-\frac{1}{2\left( u+1\right) }du.
\end{equation*}$$
$$\tag{5}$$
