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