Evaluating Indefinite Integral $$∫  \frac{1}{(1-x^3)^{1/3}}\, dx$$
I tried substituting $1-x^3$ as $t^3$ but I am not able to calculate it after that.
Thanks!
 A: For solving
$$
I=∫\frac{1}{\sqrt[3]{1-x^3}}\, \mathrm{d}x
$$
Let $t=\dfrac{x}{\sqrt[3]{1-x^3}}$,
solved
$$
\begin{aligned}
x&=\frac{t}{\sqrt[3]{t^3+1}}\\
\mathrm{d}x&=\frac{1}{(t^3+1)\sqrt[3]{t^3+1}}\mathrm{dt}\\
\end{aligned}
$$
and
$$
\frac{1}{\sqrt[3]{1-x^3}}=\sqrt[3]{t^3+1}
$$
then
$$
I = \int\frac{\sqrt[3]{t^3+1}}{(t^3+1)\sqrt[3]{t^3+1}}\mathrm{dt}=\int\frac{1}{t^3+1}\mathrm{d}t
$$

The next part is simple.
$$
\begin{aligned}
I
&= \frac{1}{3}\int\frac{1}{1+(-1)^{2/3} t}+\frac{1}{1-(-1)^{1/3}t}+\frac{1}{t+1}\mathrm{d}t\\
&= \frac{1}{3}\left(\ln (t+1)-(-1)^{1/3} \ln \left((-1)^{1/3}-t\right)+(-1)^{2/3} \ln \left(t+(-1)^{2/3}\right)\right)\\
&=\frac{1}{3} \left(\ln \left(\frac{x}{\sqrt[3]{1-x^3}}+1\right)+(-1)^{2/3} \ln \left(\frac{x}{\sqrt[3]{1-x^3}}+(-1)^{2/3}\right)-(-1)^{1/3} \ln \left((-1)^{1/3}-\frac{x}{\sqrt[3]{1-x^3}}\right)\right)
\end{aligned}
$$
Finally, take the derivation and verify that the result is correct.
D[1 / 3 (-(-1)^(1 / 3) Log[(-1)^(1 / 3) - x / (1 - x^3)^(1 / 3)] + Log[1 + x / (1 - x^3)^(1 / 3)] + (-1)^(2 / 3) Log[(-1)^(2 / 3) + x / (1 - x^3)^(1 / 3)]), x] // FullSimplify

