Answer
$$\frac{2}{7} ( x^{2}+2 x)^{7 / 4}+C$$
Work Step by Step
Given
$$ \int(x+1)\left(x^{2}+2 x\right)^{3 / 4} d x$$
Let
$$ u= x^{2}+2 x\ \ \ \Rightarrow \ \ \ du=(2x+2)dx$$
Then
\begin{aligned} \int(x+1)\left(x^{2}+2 x\right)^{3 / 4} d x &=\frac{1}{2} \int u^{3 / 4} d u \\ &=\frac{1}{2}\left(\frac{4}{7} u^{7 / 4}+C\right) \\ &=\frac{2}{7} u^{7 / 4}+C\\
&= \frac{2}{7} ( x^{2}+2 x)^{7 / 4}+C\end{aligned}
