Answer
$$\frac{2}{3}x\left| {3x + 1} \right| - \frac{1}{9}\left( {3x + 1} \right)\left| {3x + 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {2x\frac{{\left| {3x + 1} \right|}}{{3x + 1}}} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx \cr
& dv = \frac{{\left| {3x + 1} \right|}}{{3x + 1}}dx,{\text{ }}v = \int {\frac{{\left| {3x + 1} \right|}}{{3x + 1}}} dx \cr
& {\text{By the given table of integrals }} \cr
& v = \int {\frac{{\left| {3x + 1} \right|}}{{3x + 1}}} dx = \frac{1}{3}\left| {3x + 1} \right| \cr
& {\text{Using the formula of integration by parts}} \cr
& \int u dv = uv - \int v du \cr
& \int {2x\frac{{\left| {3x + 1} \right|}}{{3x + 1}}} dx = \frac{2}{3}x\left| {3x + 1} \right| - \int {\frac{1}{3}\left| {3x + 1} \right|\left( 2 \right)} dx \cr
& = \frac{2}{3}x\left| {3x + 1} \right| - \frac{2}{3}\int {\left| {3x + 1} \right|} dx \cr
& {\text{By the given table of integrals we solve }}\int {\left| {3x + 1} \right|} dx \cr
& \int {\left| {3x + 1} \right|} dx = \frac{1}{{2\left( 3 \right)}}\left( {3x + 1} \right)\left| {3x + 1} \right| + C \cr
& {\text{Therefore,}} \cr
& = \frac{2}{3}x\left| {3x + 1} \right| - \frac{2}{3}\left( {\frac{1}{{2\left( 3 \right)}}\left( {3x + 1} \right)\left| {3x + 1} \right|} \right) + C \cr
& = \frac{2}{3}x\left| {3x + 1} \right| - \frac{1}{9}\left( {3x + 1} \right)\left| {3x + 1} \right| + C \cr} $$
