Answer
$\frac{1}{2}\ln \left| {\frac{{x - 2}}{x}} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} - 2x}}}dx \cr
& {\text{Completing the square}} \cr
& \int {\frac{{dx}}{{{x^2} - 2x}}} dx = \int {\frac{{dx}}{{{x^2} - 2x + 1 - 1}}} \cr
& = \int {\frac{{dx}}{{{{\left( {x - 1} \right)}^2} - 1}}}dx \cr
& {\text{Using the formula 6: }}\int {\frac{{dx}}{{{x^2} - {a^2}}}} = \frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right| + C \cr
& \int {\frac{{dx}}{{{{\left( {x - 1} \right)}^2} - 1}}} = \frac{1}{{2\left( 1 \right)}}\ln \left| {\frac{{\left( {x - 1} \right) - 1}}{{\left( {x - 1} \right) + 1}}} \right| + C \cr
& {\text{Simplifying}} \cr
& \int {\frac{{dx}}{{{x^2} - 2x}}} = \frac{1}{2}\ln \left| {\frac{{x - 2}}{x}} \right| + C \cr} $$
