Answer
$\frac{{3{{\left( {1 - x} \right)}^{8/3}}}}{8} - \frac{{3{{\left( {1 - x} \right)}^{5/3}}}}{5} + C$
Work Step by Step
$$\eqalign{
& \int {x{{\left( {1 - x} \right)}^{2/3}}} dx \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = 1 - x,{\text{ }}x = 1 - u,{\text{ then }}dx = - du \cr
& {\text{Applying the substitution}} \cr
& \int {x{{\left( {1 - x} \right)}^{2/3}}} dx = \int {\left( {1 - u} \right){u^{2/3}}\left( { - 1} \right)} du \cr
& = \int {\left( {u - 1} \right){u^{2/3}}} du \cr
& = \int {\left( {{u^{5/3}} - {u^{2/3}}} \right)} du \cr
& {\text{Integrate }} \cr
& = \frac{{{u^{8/3}}}}{{8/3}} - \frac{{{u^{5/3}}}}{{5/3}} + C \cr
& = \frac{{3{u^{8/3}}}}{8} - \frac{{3{u^{5/3}}}}{5} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}1 - x{\text{ for }}u \cr
& = \frac{{3{{\left( {1 - x} \right)}^{8/3}}}}{8} - \frac{{3{{\left( {1 - x} \right)}^{5/3}}}}{5} + C \cr} $$
