Answer
$\frac{2}{{15}}{\left( {2 + 3{e^r}} \right)^{5/2}} + C$
Work Step by Step
$$\eqalign{
& \int {{e^r}{{\left( {2 + 3{e^r}} \right)}^{3/2}}} dr \cr
& {\text{Let }}u = 2 + 3{e^r},{\text{ then }}du = 3{e^r}dr,{\text{ }}{e^r}dr = \frac{1}{3}du \cr
& {\text{Applying the substitution}} \cr
& \int {{{\left( {2 + 3{e^r}} \right)}^{3/2}}} {e^r}dr = \int {{u^{3/2}}} \left( {\frac{1}{3}} \right)du \cr
& = \frac{1}{3}\int {{u^{3/2}}} du \cr
& {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr
& = \frac{1}{3}\left( {\frac{{{u^{5/2}}}}{{5/2}}} \right) + C \cr
& = \frac{2}{{15}}{u^{5/2}} + C \cr
& {\text{Write in terms of }}r,{\text{ substitute }}u = 2 + 3{e^r} \cr
& = \frac{2}{{15}}{\left( {2 + 3{e^r}} \right)^{5/2}} + C \cr} $$
