Answer
$$ - \frac{{3x}}{2}\left( { - x + 5} \right)\left| { - x + 5} \right| - \frac{1}{2}{\left( { - x + 5} \right)^2}\left| { - x + 5} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {3x\left| { - x + 5} \right|} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = 3x,{\text{ }}du = 3dx \cr
& dv = \left| { - x + 5} \right|dx,{\text{ }}v = \int {\left| { - x + 5} \right|} dx \cr
& {\text{By the given table of integrals}} \cr
& v = \int {\left| { - x + 5} \right|} dx = - \frac{1}{2}\left( { - x + 5} \right)\left| { - x + 5} \right| \cr
& {\text{Using the formula of integration by parts}} \cr
& \int u dv = uv - \int v du \cr
& \int {3x\left| { - x + 5} \right|} dx = - \frac{{3x}}{2}\left( { - x + 5} \right)\left| { - x + 5} \right| \cr
& {\text{ }} - \int {\left( { - \frac{1}{2}\left( { - x + 5} \right)\left| { - x + 5} \right|} \right)\left( 3 \right)} dx \cr
& = - \frac{{3x}}{2}\left( { - x + 5} \right)\left| { - x + 5} \right|{\text{ + }}\frac{3}{2}\int {\left( { - x + 5} \right)\left| { - x + 5} \right|} dx \cr
& {\text{By the given table of integrals we solve }}\int {\left( { - x + 5} \right)\left| { - x + 5} \right|} dx \cr
& \int {\left( { - x + 5} \right)\left| { - x + 5} \right|} dx = \frac{1}{{3\left( { - 1} \right)}}{\left( { - x + 5} \right)^2}\left| { - x + 5} \right| + C,{\text{ then }} \cr
& = - \frac{{3x}}{2}\left( { - x + 5} \right)\left| { - x + 5} \right|{\text{ + }}\frac{3}{2}\left( {\frac{1}{{3\left( { - 1} \right)}}{{\left( { - x + 5} \right)}^2}\left| { - x + 5} \right|} \right) + C \cr
& = - \frac{{3x}}{2}\left( { - x + 5} \right)\left| { - x + 5} \right| - \frac{1}{2}{\left( { - x + 5} \right)^2}\left| { - x + 5} \right| + C \cr} $$
