Evaluate $\int \left(x^2(1+x^3)^{\frac{2}{3}}\right)^{-1} dx$ I'm stuck on evaluating this indefinite integral.
$$\int\frac{dx}{x^2(1+x^3)^{\frac{2}{3}}}$$
I tried doing a u-substitution on the $1+x^3$ term inside the two-thirds power but didn't get anywhere. Any help?

There is a "dirty" trick having in mind that


*

*$\frac{d}{dx}\left( \sqrt[3]{1+x^3}\right)= \frac{x^2}{\left(1+x^3\right)^{\frac{2}{3}}}$
Then partial integration gives:
\begin{eqnarray*} \int\frac{dx}{x^2(1+x^3)^{\frac{2}{3}}}\; dx
& = & \int\frac{1+x^3 - x^3}{x^2\left(1+x^3\right)^{\frac{2}{3}}}\; dx \\
& = & \int\underbrace{\frac{1}{x^2}}_{u'}\underbrace{\sqrt[3]{1+x^3}}_{v}\; dx - \int\frac{x}{\left(1+x^3\right)^{\frac{2}{3}}}\; dx \\
& = & -\frac{\sqrt[3]{1+x^3}}{x} + \int\frac{1}{x}\cdot \frac{x^2}{\left(1+x^3\right)^{\frac{2}{3}}}\; dx - \int\frac{x}{\left(1+x^3\right)^{\frac{2}{3}}}\; dx \\
& = & -\frac{\sqrt[3]{1+x^3}}{x} (+C)
\end{eqnarray*}
