Answer
$\frac{2}{3}{\tan ^{ - 1}}\left( {{x^{3/2}}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt x }}{{1 + {x^3}}}} dx \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{{\sqrt x }}{{1 + {x^3}}}} dx = \int {\frac{{{x^{1/2}}}}{{1 + {{\left( {{x^{3/2}}} \right)}^2}}}} dx \cr
& {\text{Let }}u = {x^{3/2}} \to du = \frac{3}{2}{x^{1/2}}dx,{\text{ }}\frac{2}{3}du = {x^{1/2}}dx \cr
& {\text{Substituting}} \cr
& \int {\frac{{{x^{1/2}}}}{{1 + {{\left( {{x^{3/2}}} \right)}^2}}}} dx = \int {\frac{{\left( {2/3} \right)du}}{{1 + {{\left( u \right)}^2}}}} \cr
& = \frac{2}{3}\int {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{Integrating}} \cr
& = \frac{2}{3}{\tan ^{ - 1}}u + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = {x^{3/2}} \cr
& = \frac{2}{3}{\tan ^{ - 1}}\left( {{x^{3/2}}} \right) + C \cr} $$
