## Calculus (3rd Edition)

$$\frac{1}{3}\ln(x^3+2)+c.$$
By using the fact that $\int u'/u = \ln u$, we have $$\int \frac{x^2}{x^3+2}dx=\frac{1}{3}\int \frac{3x^2}{x^3+2}dx=\frac{1}{3}\ln(x^3+2)+c.$$