Answer
$$\frac{2}{3}{\tan ^{ - 1}}\left( {{x^{3/2}}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt x }}{{1 + {x^3}}}} dx \cr
& {\text{we use the hint }}u = {x^{3/2}},\,\,\,\,x = {u^{2/3}} \cr
& u = {x^{3/2}},\,\,\,\,du = \frac{3}{2}{x^{1/2}}dx,\,\,\,\,\,\sqrt x dx = \frac{2}{3}du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{\sqrt x }}{{1 + {x^3}}}} dx = \int {\frac{{\left( {2/3} \right)du}}{{1 + {{\left( {{u^{2/3}}} \right)}^3}}}} \cr
& = \frac{2}{3}\int {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{integrating}} \cr
& = \frac{2}{3}{\tan ^{ - 1}}u + C \cr
& {\text{replace }}{x^{3/2}}{\text{ for }}u \cr
& = \frac{2}{3}{\tan ^{ - 1}}\left( {{x^{3/2}}} \right) + C \cr} $$
