How to solve $\int\frac{dx}{\sqrt[3]{1+x^3}}$?

Question

I have:

$$\int\frac{dx}{\sqrt[3]{1+x^3}}$$

It seems easy but neither $t^3=1+x^3$ nor $t = \sqrt[3]{1+x^3}$ nor $t = x^3$ made integral easier. What is the way to solve it then?

Answer 1

This integral in fact has elementary antiderivative.

Let $x=1/u$, $dx=-du/u^2 $, we have $$\int \frac{1}{\sqrt[3]{1+x^3}}dx = -\int\frac{1}{u^2\sqrt[3]{1+\frac{1}{u^3}}} du = -\int\frac{u^2}{u^3 \sqrt[3]{1+u^3}} du = -\frac{1}{3}\int\frac{d(u^3)}{u^3 \sqrt[3]{1+u^3}}$$

Letting $1+u^3 = t^3$, the original integral is transformed into $$-\int\frac{t}{t^3-1} dt$$ from which I believe you can continue.