Answer
$$2x + 6\ln \left| {x - 3} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2x}}{{x - 3}}} dx \cr
& {\text{Let }}u = x - 3,{\text{ }}x = u + 3,{\text{ }}dx = du \cr
& {\text{Write in terms of }}u \cr
& \int {\frac{{2x}}{{x - 3}}} dx = \int {\frac{{2\left( {u + 3} \right)}}{u}} du \cr
& = \int {\frac{{2u + 6}}{u}} du \cr
& = \int {\left( {\frac{{2u}}{u} + \frac{6}{u}} \right)} du \cr
& = \int {\left( {2 + \frac{6}{u}} \right)} du \cr
& {\text{Integrate}} \cr
& = 2u + 6\ln \left| u \right| + C \cr
& {\text{Write in terms of }}u \cr
& = 2\left( {x - 3} \right) + 6\ln \left| {x - 3} \right| + C \cr
& {\text{Simplify and combine constants}} \cr
& = 2x - 6 + 6\ln \left| {x - 3} \right| + C \cr
& = 2x + 6\ln \left| {x - 3} \right| + C \cr} $$
