Answer
$$ - {x^2}\left( { - x + 4} \right)\left| { - x + 4} \right| - \frac{2}{3}x{\left( { - x + 4} \right)^2}\left| { - x + 4} \right| - \frac{1}{6}{\left( { - x + 4} \right)^3}\left| { - x + 4} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {2{x^2}\left| { - x + 4} \right|} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = 2{x^2},{\text{ }}du = 4xdx \cr
& dv = \left| { - x + 4} \right|dx,{\text{ }}v = \int {\left| { - x + 4} \right|} dx \cr
& {\text{By the given table of integrals }}\int {\left| {ax + b} \right| = \frac{1}{{2a}}\left( {ax + b} \right)\left| {ax + b} \right| + C} \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 {2{x^2}\left| { - x + 4} \right|} dx = - \frac{{2{x^2}}}{2}\left( { - x + 4} \right)\left| { - x + 4} \right| \cr
& {\text{ }} + \int {\frac{1}{2}\left( { - x + 4} \right)\left| { - x + 4} \right|\left( {4x} \right)} dx{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{Integrating }}\int {\frac{1}{2}\left( { - x + 4} \right)\left| { - x + 4} \right|\left( {4x} \right)} dx{\text{ by parts}} \cr
& = \int {2x\left( { - x + 4} \right)\left| { - x + 4} \right|} dx \cr
& = \int {2x\left( { - x + 4} \right)\left| { - x + 4} \right|} dx \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx \cr
& dv = \left( { - x + 4} \right)\left| { - x + 4} \right|dx,{\text{ }}v = - \frac{1}{3}{\left( { - x + 4} \right)^2}\left| { - x + 4} \right| \cr
& {\text{Using the formula of integration by parts}} \cr
& \int u dv = uv - \int v du \cr
& = - \frac{{2x}}{3}{\left( { - x + 4} \right)^2}\left| { - x + 4} \right| + \int {\frac{1}{3}{{\left( { - x + 4} \right)}^2}\left| { - x + 4} \right|} \left( 2 \right)dx \cr
& = - \frac{{2x}}{3}{\left( { - x + 4} \right)^2}\left| { - x + 4} \right| + \frac{2}{3}\int {{{\left( { - x + 4} \right)}^2}\left| { - x + 4} \right|} dx \cr
& {\text{By the given tables }} \cr
& \int {{{\left( {ax + b} \right)}^2}\left| {ax + b} \right|dx = \frac{1}{{4a}}{{\left( {ax + b} \right)}^3}} \left| {ax + b} \right| + C,{\text{ then}} \cr
& = - \frac{{2x}}{3}{\left( { - x + 4} \right)^2}\left| { - x + 4} \right| + \frac{2}{3}\left( { - \frac{1}{4}{{\left( { - x + 4} \right)}^3}\left| { - x + 4} \right|} \right) + C \cr
& = - \frac{{2x}}{3}{\left( { - x + 4} \right)^2}\left| { - x + 4} \right| - \frac{1}{6}{\left( { - x + 4} \right)^3}\left| { - x + 4} \right| + C \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& = - \frac{{2{x^2}}}{2}\left( { - x + 4} \right)\left| { - x + 4} \right| - \frac{{2x}}{3}{\left( { - x + 4} \right)^2}\left| { - x + 4} \right| \cr
& - \frac{1}{6}{\left( { - x + 4} \right)^3}\left| { - x + 4} \right| + C \cr
& {\text{Simplifying}} \cr
& = - {x^2}\left( { - x + 4} \right)\left| { - x + 4} \right| - \frac{2}{3}x{\left( { - x + 4} \right)^2}\left| { - x + 4} \right| \cr
& - \frac{1}{6}{\left( { - x + 4} \right)^3}\left| { - x + 4} \right| + C \cr} $$
