TITLE: Evaluate : $\int_0^1 \frac{dx}{\sqrt[3]{1-x^3}}$.
QUESTION [6 upvotes]: Evaluate : $$ \int_0^1 \frac{\mathrm{d} x}{\sqrt[3]{1-x^3}}.$$

if you feel that this integral is easy, just post hints.

REPLY [2 votes]: You do not need any beta- or gamma-functions for this integral. I will deal with indefinite integral, you should take care of the limits.
Use the substitution 
$$
1-\frac{1}{x^3}=t^3.
$$
Then, you have
$$
x=\frac{1}{\sqrt[3]{1-t^3}},\quad dx=\frac{t^2\,dt}{(1-t^3)^{4/3}},\quad \sqrt[3]{1-x^3}=-\frac{t}{\sqrt[3]{1-t^3}}.
$$
Plugging everything in, you get
$$
-\int \frac{t}{1-t^3}dt,
$$
which can be integrated using standard methods (e.g., partial fractions).
The big question is how would I know about this substitution? You can find a recipe in this post.