Answer
$$
\int \frac{ x^2d t}{x^3+2}= \frac{1}{3}\ln |x^3+2|+c.
$$
Work Step by Step
Le $ u=x^3+2$ and hence $ du=3x^2dx $, then we have
$$
\int \frac{ x^2d t}{x^3+2}=\frac{1}{3}\int \frac{ d u}{u}= \frac{1}{3}\ln u+c=\frac{1}{3}\ln |x^3+2|+c.
$$
