How do we integrate, $\int \frac{1}{x+\frac{1}{x^2}}dx$? How do we integrate the following integral?
$$\int \frac{1}{x+\large\frac{1}{x^2}}\,dx\quad\text{where}\;\;x\ne-1$$

The integral is equivalent to
$$\int dx \frac{x^2}{1+x^3} = \frac{1}{3} \int \frac{d(x^3)}{1+x^3}  = \frac{1}{3} \log{(1+x^3)} + C$$
