Evaluate $\int \left(x^2(1+x^3)^{\frac{2}{3}}\right)^{-1} dx$

Question

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?

Answer 1

Hint:

Instead, use the substitution $u=\dfrac{1}{x^3} + 1$, $du = -\dfrac{3}{x^4} dx$.

Answer 2

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*}