Evaluating Indefinite Integral

Question

$$∫ \frac{1}{(1-x^3)^{1/3}}\, dx$$

I tried substituting $1-x^3$ as $t^3$ but I am not able to calculate it after that.
Thanks!

Answer 1

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

$$ I = \int\frac{\sqrt[3]{t^3+1}}{(t^3+1)\sqrt[3]{t^3+1}}\mathrm{dt}=\int\frac{1}{t^3+1}\mathrm{d}t $$

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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