Answer
$$
\int x^2 \sqrt{x+1} dx
=\frac{2}{7}(x+1)^{7/2} -\frac{4}{5}(x+1)^{5/2} +\frac{2}{3}(x+1)^{3/2} +c
$$
Work Step by Step
Since $ u=x+1$, then $ du=dx $ and hence, $$
\int x^2 \sqrt{x+1} dx=\int (u-1)^2u^{1/2} d u= \int (u^2-2u+1)u^{1/2} d u\\
=\int u^{5/2} -2u^{3/2} +u^{1/2} d u=\frac{2}{7}u^{7/2} -2\frac{2}{5}u^{5/2} +\frac{2}{3}u^{3/2} +c\\
=\frac{2}{7}(x+1)^{7/2} -\frac{4}{5}(x+1)^{5/2} +\frac{2}{3} (x+1)^{3/2} +c
$$
