Answer
$$
\int t\sqrt{t^2+1} dt
=\frac{1}{3} (t^2+1)^{3/2}+c.
$$
Work Step by Step
Since $ u=t^2+1 $, then $ du=2tdt $ and hence, $$
\int t\sqrt{t^2+1} dt=\frac{1}{2}\int u^{1/2} d u=\frac{1}{2} \frac{2}{3}u^{3/2}+c\\
=\frac{1}{3} (t^2+1)^{3/2}+c.
$$
