Answer
\[\frac{1}{2}\ln\left|\frac{x-2}{x}\right|+C\]
Work Step by Step
Let \[I=\int\frac{dx}{x^2-2x}=\frac{dx}{(x-1)^2-1}\]
Substitute $t=x-1$ _____(1)
$\;\;\;\;\;\;\;\;\;\Rightarrow dt=dx$
\[I=\int\frac{dt}{t^2-1}\]
\[\left[\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|\right]\]
\[I=\frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|+C\]
$C$ is constant of integration
From (1)
\[I=\frac{1}{2}\ln\left|\frac{x-2}{x}\right|+C\]
Hence $I=\large\frac{1}{2}\ln\left|\frac{x-2}{x}\right|$ $+C$.
