Convergence of $\displaystyle\int\frac{1}{\sqrt[3]{1-x^3}}\ dx$

Question

Please help me to prove that this integral converges.

$$\int_{0}^1 \frac{1}{\sqrt[3]{1-x^3}}\ dx $$

No ideas. Tried to find function which is bigger and converges, equivalent fun-s, but no result still.

Answer 1

Using the factorization $a^3-b^3 = (a-b)(a^2+ab +b^2) $, rewrite the integral as

$$\int_0^1 dx \, \frac{(1-x)^{-1/3}}{(1+x+x^2)^{1/3}} $$

Sub $y=1-x$ and observe that

$$\int dy \, y^{-1/3} = \frac{3}{2} y^{2/3} + C$$

Answer 2

Hint: Use the fact that $\sqrt{1-x^2}<\sqrt[3]{1-x^3}<1$ for $x\in(0,1)$.