Answer
$$ \frac{x}{2}\left( {2x + 1} \right)\left| {2x + 1} \right| - \frac{1}{{12}}{\left( {2x + 1} \right)^2}\left| {2x + 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {2x\left| {2x + 1} \right|} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx \cr
& dv = \left| {2x + 1} \right|dx,{\text{ }}v = \int {\left| {2x + 1} \right|} dx \cr
& {\text{By the given table of integrals}} \cr
& v = \int {\left| {2x + 1} \right|} dx = \frac{1}{{2\left( 2 \right)}}\left( {2x + 1} \right)\left| {2x + 1} \right| \cr
& v = \frac{1}{4}\left( {2x + 1} \right)\left| {2x + 1} \right| \cr
& {\text{Using the formula of integration by parts}} \cr
& \int u dv = uv - \int v du \cr
& \int {2x\left| {2x + 1} \right|} dx = \frac{x}{2}\left( {2x + 1} \right)\left| {2x + 1} \right| - \frac{1}{2}\int {\left( {2x + 1} \right)\left| {2x + 1} \right|} dx \cr
& {\text{By the given table of integrals we solve }}\int {\left( {2x + 1} \right)\left| {2x + 1} \right|} dx \cr
& \int {\left( {2x + 1} \right)\left| {2x + 1} \right|} dx = \frac{1}{{3\left( 2 \right)}}{\left( {2x + 1} \right)^2}\left| {2x + 1} \right| + C,{\text{ then }} \cr
& = \frac{x}{2}\left( {2x + 1} \right)\left| {2x + 1} \right| - \frac{1}{2}\left( {\frac{1}{{3\left( 2 \right)}}{{\left( {2x + 1} \right)}^2}\left| {2x + 1} \right|} \right) + C \cr
& = \frac{x}{2}\left( {2x + 1} \right)\left| {2x + 1} \right| - \frac{1}{{12}}{\left( {2x + 1} \right)^2}\left| {2x + 1} \right| + C \cr} $$
