Answer
$$
\int \frac{t^{2}}{\sqrt{1-t^{6}}} d t =\frac{1}{3} \sin ^{-1}\left(t^{3}\right)+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\int \frac{t^{2}}{\sqrt{1-t^{6}}} d t
$$
Let $u=t^{3} .$ Then $d u=3 t^{2} d t $. and, the Substitution Rule gives
$$
\begin{aligned}
\int \frac{t^{2}}{\sqrt{1-t^{6}}} d t& =\int \frac{\frac{1}{3} d u}{\sqrt{1-u^{2}}} \\
&=\frac{1}{3} \sin ^{-1} u+C \\
&=\frac{1}{3} \sin ^{-1}\left(t^{3}\right)+C
\end{aligned}
$$
where $C$ is an arbitrary constant.