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