Answer
$$\int \frac{\sqrt{x}}{1+x^{3}}dx=\frac{2}{3}arctan\,(x^{\frac{3}{2}})+C$$
Work Step by Step
$$let\ t=x^{\frac{3}{2}},dt=\frac{3}{2}\sqrt{x}dx$$
$$\int \frac{\sqrt{x}}{1+x^{3}}dx=\frac{2}{3}\int \frac{1}{1+t^{2}}dt$$
$$=\frac{2}{3}arctan\,t+C=\frac{2}{3}arctan\,(x^{\frac{3}{2}})+C$$
