Evaluating $\int \tan^{1/3}(\theta) d \theta$ I know $\int \tan^{1/3}\theta  d \theta$ is a non integrable but wolfram alpha does it easily. And can anyone explain how a fuction is non integrable? My argument is that there has to be a function which represents the area under a graph. It's just that we do not know it. Please shed a bit light on it. Also, how can we solve $\int \tan^{1/3}\theta  d \theta$?

By enforcing the substitutions $x=\arctan(t)$, then $t=u^3$, we get:
$$ \int\left(\tan x\right)^{1/3}\,dx = \int\frac{t^{1/3} dt}{1+t^2}=\int\frac{3u^3}{1+u^6}\,du $$
and the last integral can be computed through a partial fraction decomposition. We have:
$$ \int\frac{3u^3}{1+u^6}\,du =C+\frac{\sqrt{3}}{2}\arctan\left(\frac{2u^2-1}{\sqrt{3}}\right)+\frac{1}{4}\log\left(\frac{1+u^6}{(1+u^2)^3}\right),$$
not that simple.
