Answer
$$\frac{2}{7}(t-3)^{7/2}+\frac{12}{5}(t-3)^{5/2}+6(t-3)^{3/2}+c.$$
Work Step by Step
Let $u=t-3$, then $du= dt$ and hence we get
$$\int t^2\sqrt{t-3}dt= \int (u+3)^2u^{1/2}du \\
=\int (u^{5/2}+6u^{3/2}+9u^{1/2})du\\
=\frac{1}{7/2}u^{7/2}+\frac{6}{5/2}u^{5/2}+\frac{9}{3/2}u^{3/2}+c\\
=\frac{2}{7}u^{7/2}+\frac{12}{5}u^{5/2}+\frac{18}{3}u^{3/2}+c\\
=\frac{2}{7}(t-3)^{7/2}+\frac{12}{5}(t-3)^{5/2}+6(t-3)^{3/2}+c.$$
