Answer
$$\displaystyle \int{\frac{\sqrt{x}}{1+x^3}dx}=\frac{2}{3}\arctan \left( x^{3/2} \right) +C$$
Work Step by Step
$\displaystyle I=\int{\frac{\sqrt{x}}{1+x^3}dx}$
$\displaystyle{\left[\begin{matrix}
u=x^{\frac{3}{2}}& dx=\frac{2du}{3\sqrt{x}}& u^2=x^3\\
\end{matrix}\right]}$
$\displaystyle{I=\int{\frac{\sqrt{x}}{1+u^2}\times \frac{2du}{3\sqrt{x}}}\\
I=\frac{2}{3}\int{\frac{1}{1+u^2}du}\\
I=\frac{2}{3}\arctan u+C\\
I=\frac{2}{3}\arctan \left( x^{3/2} \right) +C}$
