Answer
$$ \frac{3}{7}(x+c)^{7/3}- \frac{3c}{4}(x+c)^{4/3}+c_1$$
Work Step by Step
Given
\begin{aligned}
\int x\sqrt[3]{x+c}dx
\end{aligned}
Let
$$u^3=x+c\ \ \ \to \ \ 3u^2du=dx$$
then
\begin{aligned}
\int x\sqrt[3]{x+c}dx &= \int (u^3-c) (u)(3u^2)du\\
&=\int (3u^6- 3cu^3)du\\
&=\frac{3}{7}u^7- \frac{3c}{4}u^4+c_1\\
&= \frac{3}{7}(x+c)^{7/3}- \frac{3c}{4}(x+c)^{4/3}+c_1
\end{aligned}