Integrate $\int \frac{x}{(1-x^6)^{1/3}}\ dx$ My friend challenged me to integrate $\tan^{1/3}(t)$ without using the normal method of just letting $\tan(t) = x^3$ and proceeding...
I tried a different approach
$$\int\tan^{1/3}(t) dt = \int \frac{\sin^{1/3} t }{\cos^{1/3} t} dt$$
Putting $\cos t = x^3$ gives
$$\int \frac{x}{(1-x^6)^{1/3}} dx = \frac12 \int \frac{dy}{(1-y^3)^{1/3}}$$
where $y = x^2$.
Is there any way to solve it further?
 A: Continue with the substitution $y=\frac t{(1+t^3)^{1/3}}$
\begin{align}
& \int \frac{1}{(1-y^3)^{1/3}}dy\\
=&\int \frac1{1+t^3}dt= \frac16\ln\frac{(1+t)^2}{1-t+t^2}+\frac1{\sqrt3}\tan^{-1}\frac{2t-1}{\sqrt3}+C
\end{align}
A: Letting $y^3=\frac 1{x^6}-1$, transforms the integral into
$$
I=-\frac{1}{2} \int \frac{y d y}{y^3+1}
$$
Resolving the integrand into partial fractions
$$\frac{y}{y^3+1}= -\frac{1}{3(y+1)}+\frac{y+1}{3\left(y^2-y+1\right)}$$
we have
$$
\begin{aligned}
 \int \frac{y d y}{y^3+1} =& \int\left(-\frac{1}{3(y+1)}+\frac{y+1}{3\left(y^2-y+1\right)}\right) d y \\
=&-\frac{1}{3} \ln |y+1|+\frac{1}{6} \int \frac{(2 y-1)+3}{y^2-y+1} d y \\
=&-\frac{1}{3} \ln |y+1|+\frac{1}{6} \ln \left|y^2-y+1\right|+2 \int \frac{d y}{(2 y-1)^2+3} \\
=& \frac{1}{6}\left(\ln\frac{\left| y^2-y+1\right|}{(y+1)^2}+4 \sqrt{3} \tan ^{-1} \frac{2 y-1}{\sqrt{3}}\right)+c_1
\end{aligned}
$$
$$
I=\frac{1}{12}\left(\ln \frac{ (y+1)^2}{ \left|y^2-y+1\right|}-4 \sqrt{3} \tan ^{-1} \frac{2 y-1}{\sqrt{3}}\right)+C
$$
where $y^3=\frac 1{x^6}-1$.
