Answer
$$2x\left| {x - 3} \right| - \left( {x - 3} \right)\left| {x - 3} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {2x\frac{{\left| {x - 3} \right|}}{{x - 3}}} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx \cr
& dv = \frac{{\left| {x - 3} \right|}}{{x - 3}}dx,{\text{ }}v = \int {\frac{{\left| {x - 3} \right|}}{{x - 3}}} dx \cr
& {\text{By the given table of integrals }} \cr
& v = \int {\frac{{\left| {x - 3} \right|}}{{x - 3}}} dx = \left| {x - 3} \right| \cr
& {\text{Using the formula of integration by parts}} \cr
& \int u dv = uv - \int v du \cr
& \int {2x\frac{{\left| {x - 3} \right|}}{{x - 3}}} dx = 2x\left| {x - 3} \right| - \int {\left| {x - 3} \right|\left( 2 \right)} dx \cr
& = 2x\left| {x - 3} \right| - 2\int {\left| {x - 3} \right|} dx \cr
& {\text{By the given table of integrals we solve }}\int {\left| {x - 3} \right|} dx \cr
& = 2x\left| {x - 3} \right| - 2\left( {\frac{1}{2}\left( {x - 3} \right)\left| {x - 3} \right|} \right) + C \cr
& = 2x\left| {x - 3} \right| - \left( {x - 3} \right)\left| {x - 3} \right| + C \cr} $$
