## Calculus (3rd Edition)

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