Answer
$\frac{2}{7}(t+8)^{7/2}-\frac{32}{5}(t+8)^{5/2}+\frac{128}{3}(t+8)^{3/2}+c.$
Work Step by Step
Let $u=t+8$, then $du=dt$. Now, we have
\begin{align*}
\int t^2(t+8)^{1/2} dt&=\int (u-8)^2u^{1/2} du\\
&=\int u^{5/2} -16u^{3/2}+64 u^{1/2}du\\
&=\frac{1}{7/2}u^{7/2}-\frac{16}{5/2}u^{5/2}+\frac{64}{3/2}u^{3/2}+c\\
&=\frac{2}{7}(t+8)^{7/2}-\frac{32}{5}(t+8)^{5/2}+\frac{128}{3}(t+8)^{3/2}+c.
\end{align*}
