Answer
\[x + 2\ln \,\left( {x - 2} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{x}{{x - 2}}dx} \hfill \\
\hfill \\
set\,\,the\,\,substitution \hfill \\
\hfill \\
u = x - 2\,\,\,\,\,\,then\,\,\,\,\,du = dx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{x}{{x - 2}}dx} = \int_{}^{} {\frac{{u + 2}}{u}du} \hfill \\
\hfill \\
= \int_{}^{} {1 + \frac{2}{u}du} \hfill \\
\int_{}^{} {1dx} \,\left( {since\,\,du = dx} \right)\,\,2\int_{}^{} {\frac{1}{u}du} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
x + 2\ln \left| u \right| + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,u = x - 2 \hfill \\
\hfill \\
x + 2\ln \,\left( {x - 2} \right) + C \hfill \\
\hfill \\
\end{gathered} \]
