Answer
$\frac{{2{{\left( {x + 2} \right)}^{5/2}}}}{5} - \frac{{4{{\left( {x + 2} \right)}^{3/2}}}}{3} + C$
Work Step by Step
$$\eqalign{
& \int {x\sqrt {x + 2} } dx \cr
& {\text{Let }}u = x + 2 \to x = u - 2,{\text{ }}dx = du \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& \int {x\sqrt {x + 2} } dx = \int {\left( {u - 2} \right)\sqrt u } du \cr
& = \int {\left( {u - 2} \right){u^{1/2}}} du \cr
& = \int {\left( {{u^{3/2}} - 2{u^{1/2}}} \right)} du \cr
& {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr
& = \frac{{{u^{5/2}}}}{{5/2}} - 2\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \cr
& = \frac{{2{u^{5/2}}}}{5} - \frac{{4{u^{3/2}}}}{3} + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}u = x + 2 \cr
& = \frac{{2{{\left( {x + 2} \right)}^{5/2}}}}{5} - \frac{{4{{\left( {x + 2} \right)}^{3/2}}}}{3} + C \cr} $$