Integrate:
$\displaystyle \text{I}=\int \sqrt[3]{\tan x}$
Step 1: I'd start by making the substitution $\displaystyle w^3 = \tan x \Rightarrow 3 w^2 dw = \sec^2 x dx \Rightarrow dx = \frac{3 w^2 \, dw}{\sec^2 x} = \frac{3 w^2 \, dw}{1 + \tan^2 x} = \frac{3 w^2 \, dw}{w^6 + 1}$.
Step 2: Integrate $\displaystyle \int \frac{3 w^3}{w^6 + 1} \, dw$.
Try it from here.
Hint: Use a partial fraction decomposition. Note:
$\displaystyle w^6 + 1 = (w^2)^3 + 1^3 = (w^2 + 1)(w^4 - w^2 + 1)$
$\displaystyle w^4 - w^2 + 1 = (w^2 + 1)^2 - 3w^2 = (w^2 + 1 - \sqrt{3} w)(w^2 + 1 + \sqrt{3} w)$.
Therefore: $\displaystyle w^6 + 1 = (w^2 + 1)(w^2 - \sqrt{3} w + 1)(w^2 + \sqrt{3} w + 1)$.
Following up on this, put $\displaystyle w=\frac1z$ and the integral becomes $\displaystyle \int{\frac{w^{3}}{w^{6}+1}\,dw}=-\int{\frac{z}{z^{6}+1}\,dz}=-\int{\frac{z}{\left( z^{2} \right)^{3}+1}\,dz},$ now put $\displaystyle u=z^2$ and the partial fraction descomposition becomes a bit easier.