Answer
$$6\left(\frac{1}{7}x^{7/6}-\frac{1}{5}x^{5/6}+\frac{1}{3}x^{1/2}-1+\tan^{-1}x^{1/6}\right)+C$$
Work Step by Step
Given $$\int \frac{x^{1 / 2} d x}{x^{1 / 3}+1}$$
Let $$u^6 =x\ \ \ \ \ \ 6u^5= dx $$
Then
\begin{align*}
\int \frac{x^{1 / 2} d x}{x^{1 / 3}+1}&=\int \frac{6u^8 d u}{u^2+1}\\
&= 6\int \left(u^6-u^4+u^2-1+\frac{1}{u^2+1}\right)du\\
&=6\left(\frac{1}{7}u^7-\frac{1}{5}u^5+\frac{1}{3}u^3-1+\tan^{-1}u\right)+C\\
&= 6\left(\frac{1}{7}x^{7/6}-\frac{1}{5}x^{5/6}+\frac{1}{3}x^{1/2}-1+\tan^{-1}x^{1/6}\right)+C
\end{align*}
