Answer
$$\frac{1}{2}x\left( {x + 4} \right)\left| {x + 4} \right| - \frac{1}{6}{\left( {x + 4} \right)^2}\left| {x + 4} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {x\left| {x + 4} \right|} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = x,{\text{ }}du = dx \cr
& dv = \left| {x + 4} \right|dx,{\text{ }}v = \int {\left| {x + 4} \right|} dx \cr
& {\text{By the given table of integrals }} \cr
& v = \int {\left| {x + 4} \right|} dx = \frac{1}{2}\left( {x + 4} \right)\left| {x + 4} \right| \cr
& {\text{Using the formula of integration by parts}} \cr
& \int u dv = uv - \int v du \cr
& \int {x\left| {x + 4} \right|} dx = \frac{1}{2}x\left( {x + 4} \right)\left| {x + 4} \right| - \int {\frac{1}{2}\left( {x + 4} \right)\left| {x + 4} \right|} dx \cr
& \int {x\left| {x + 4} \right|} dx = \frac{1}{2}x\left( {x + 4} \right)\left| {x + 4} \right| - \frac{1}{2}\int {\left( {x + 4} \right)\left| {x + 4} \right|} dx \cr
& {\text{By the given table of integrals we solve }}\int {\left( {x + 4} \right)\left| {x + 4} \right|} dx \cr
& \int {\left( {x + 4} \right)\left| {x + 4} \right|} dx = \frac{1}{3}{\left( {x + 4} \right)^2}\left| {x + 4} \right| + C,{\text{ then }} \cr
& \int {x\left| {x + 4} \right|} dx = \frac{1}{2}x\left( {x + 4} \right)\left| {x + 4} \right| - \frac{1}{6}{\left( {x + 4} \right)^2}\left| {x + 4} \right| + C \cr} $$
