Answer
$$\frac{2}{3}\tan^{-1}(x^{3/2})+C $$
Work Step by Step
Given $$\int \frac{\sqrt{x}}{x^{3}+1} d x$$
Let $$u=\sqrt{x},\ \ \ \ du=\frac{1}{2\sqrt{x}}dx $$
Then
\begin{aligned}
\int \frac{\sqrt{x}}{x^{3}+1} d x=& \int \frac{u}{u^{6}+1} 2 u \, d u \\
&=\int \frac{2 u^{2}}{u^{6}+1} d u \\
&=\int \frac{2 u^{2}}{\left(u^{3}\right)^{2}+1}\\
&= \frac{2}{3}\int \frac{3 u^{2}}{\left(u^{3}\right)^{2}+1}\\
&=\frac{2}{3}\tan^{-1}(u^3)+C\\
&=\frac{2}{3}\tan^{-1}(x^{3/2})+C
\end{aligned}
