Answer
$x + 3\ln \left| {x - 3} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{x - 3}}} dx \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = x - 3,{\text{ }}x = u + 3,{\text{ then }}dx = du \cr
& {\text{Applying the substitution}} \cr
& \int {\frac{x}{{x - 3}}} dx = \int {\frac{{u + 3}}{u}} du \cr
& {\text{Distribute the numerator}} \cr
& = \int {\left( {\frac{u}{u} + \frac{3}{u}} \right)} du \cr
& = \int {\left( {1 + \frac{3}{u}} \right)} du \cr
& = \int {du} + \int {\frac{3}{u}} du \cr
& {\text{Integrate }} \cr
& = u + 3\ln \left| u \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}x - 3{\text{ for }}u \cr
& = \left( {x - 3} \right) + 3\ln \left| {x - 3} \right| + C \cr
& {\text{Combining constants}} \cr
& = x + 3\ln \left| {x - 3} \right| + C \cr} $$