Answer
$${\left( {x + 4} \right)^{3/2}}\left( {\frac{{2{{\left( {x + 4} \right)}^2}}}{7} - \frac{{16\left( {x + 4} \right)}}{5} + \frac{{32}}{3}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}\sqrt {x + 4} } dx \cr
& {\text{integrate by substitution}} \cr
& {\text{set }}u = x + 4{\text{ then }}x = u - 4,\,\,\,\,\,\,\,\,dx = du \cr
& {\text{write the integral in terms of }}u \cr
& \int {{x^2}\sqrt {x + 4} } dx = \int {{{\left( {u - 4} \right)}^2}\sqrt u } du \cr
& {\text{write }}\sqrt u {\text{ as }}{u^{1/2}} \cr
& = \int {{{\left( {u - 4} \right)}^2}{u^{1/2}}} du \cr
& {\text{expand }}{\left( {u - 4} \right)^2} \cr
& = \int {\left( {{u^2} - 8u + 16} \right){u^{1/2}}} du \cr
& = \int {\left( {{u^{5/2}} - 8{u^{3/2}} + 16{u^{1/2}}} \right)} du \cr
& {\text{integrate by using the power rule}} \cr
& = \frac{{{u^{7/2}}}}{{7/2}} - 8\left( {\frac{{{u^{5/2}}}}{{5/2}}} \right) + 16\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \cr
& = \frac{{2{u^{7/2}}}}{7} - \frac{{16{u^{5/2}}}}{5} + \frac{{32{u^{3/2}}}}{3} + C \cr
& {\text{Factor }} \cr
& = {u^{3/2}}\left( {\frac{{2{u^2}}}{7} - \frac{{16u}}{5} + \frac{{32}}{3}} \right) + C \cr
& {\text{replace }}x + 4{\text{ for }}u \cr
& = {\left( {x + 4} \right)^{3/2}}\left( {\frac{{2{{\left( {x + 4} \right)}^2}}}{7} - \frac{{16\left( {x + 4} \right)}}{5} + \frac{{32}}{3}} \right) + C \cr} $$
