Answer
$\frac{66838}{105}$
Work Step by Step
Let $u=t+8$, then $du=dt$ and $u:9\to16$. Now, we have
\begin{align*}
\int_1^8 t^2(t+8)^{1/2} dt& =\int_{9}^{16} (u-8)^2u^{1/2} du\\
& =\int_{9}^{16} 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}|_{9}^{16} \\
& =\frac{2}{7}(16^{7/2}-9^{7/2})-\frac{32}{5}(16^{5/2}-9^{5/2})+\frac{128}{3}(16^{3/2}-9^{3/2})\\
&=\frac{66838}{105}.
\end{align*}
