Integration involving beta and hypergeometric functions/ differential binomial So, hey, everybody!
I have to integrate this 
$$
\int_0^2 \sqrt[3]{\frac{x^2}{2-x}} \, dx 
$$
and I've already figured out that due to Chebyshev's theorem it cannot be done in terms of elementary functions, since we can rewrite the task as 
$$ \int_0^2 x^{\frac{2}{3}}\left(2-x\right)^{-\frac{1}{3}} \, dx $$ and 
$$ \frac{2}{3}+1-\frac{1}{3}=\frac{4}{3}$$
which is, obviously, not an integer number. 
But, the point is that I'm not quite familiar with Beta-function and stuff like that, so, how do you write the answer to this problem? 
Mathematica show something like that $$\frac{\Gamma\left[\frac{1}{2} \right]\Gamma\left[\frac{2}{3} \right]  }{\Gamma\left[\frac{7}{6} \right] } $$
The question is: "Hey, how did Mathematica do this?"
 A: Here's a way to solve by using beta and gamma functions.
Let 
$$I=\int_0^2 \frac{x^{2/3}}{(2-x)^{1/3}}\,dx$$
The above is equivalent to:
$$I=\int_0^2 \frac{(2-x)^{2/3}}{x^{1/3}}\,dx$$
Add the two expressions for $I$ to obtain:
$$2I=\int_0^2 \frac{2}{x^{1/3}(2-x)^{1/3}}\,dx \Rightarrow I=\int_0^2\frac{1}{x^{1/3}(2-x)^{1/3}}\,dx=2\int_0^1 \frac{dx}{x^{1/3}(2-x)^{1/3}}$$
$$\Rightarrow I=2\int_0^1 \frac{dx}{(1-x)^{1/3}(1+x)^{1/3}}=2\int_0^1 (1-x^2)^{-1/3}\,dx$$
Use the substitution $x^2=t \Rightarrow dx=\frac{1}{2}t^{-1/2}\,dt$ to get:
$$I=\int_0^1 t^{-1/2}(1-t)^{-1/3}=B\left(\frac{1}{2},\frac{2}{3}\right)=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{7}{6}\right)}$$
You can even simplify it more by writing $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ using Euler's reflection formula.
In a similar way, you can prove the results presented by Claude Leibovici.
A: It is relative easy with the change of variable $x=2t$ and the formulas for the Beta function from  http://en.wikipedia.org/wiki/Beta_function
$$ \int_0^2 x^{\frac{2}{3}}\left(2-x\right)^{-\frac{1}{3}} \, dx 
= \int_0^1 (2t)^{\frac{2}{3}}\left(2-2t\right)^{-\frac{1}{3}} 2\, dt
= 2^{1+\frac{2}{3}-\frac{1}{3}} \int_0^1 t^{\frac{2}{3}}\left(1-t\right)^{-
\frac{1}{3}} \, dt
=2^{\frac{4}{3}} B\left(\frac{5}{3},\frac{2}{3}\right)$$
The last expression is the same as the Mathematica $\Gamma$ term.
A: I am not sure that this is the answer you expect but CAS are able to compute the antiderivative and find $$\int \sqrt[3]{\frac{x^2}{2-x}} \, dx=\frac{3}{4} \sqrt[3]{\frac{x^2}{2-x}} \left(2^{2/3} \sqrt[3]{2-x} \,
   _2F_1\left(\frac{1}{3},\frac{2}{3};\frac{5}{3};\frac{x}{2}\right)+x-2\right)$$ If we compute the integral $$I(a)=\int_0^a \sqrt[3]{\frac{x^2}{2-x}} \, dx$$ there are very few cases where the result is quite nice. The only ones I found are $$I(1)=\frac{3}{4} \left(\frac{\sqrt{\pi } \Gamma \left(\frac{5}{3}\right)}{\Gamma
   \left(\frac{7}{6}\right)}-1\right)$$ $$I(2)=\frac{\sqrt{\pi } \Gamma \left(\frac{2}{3}\right)}{\Gamma \left(\frac{7}{6}\right)}$$
