<s>[INST] Calculate integral using beta and gamma functions I have to calculate the following integral using beta and gamma functions:
$$
\int\limits_0^1 \frac{x\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}}
$$
I came up with this terrible solution. Firstly, let's break it into two parts:
$$
\int\limits_0^1 \frac{(x-2)\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}} + \int\limits_0^1 \frac{2\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}}
$$
The first one is $-B\left(\frac 13,\frac 23\right)$. The second one can be simplified with substitution $x = 1 - \frac 1t$ to
$$
2\int\limits_{-\infty}^0 \frac{dt}{(t+1)(t-1)^\frac 23}
$$
But it's too unwieldy in my opinion. Furthermore, it's not so easy to evaluate $B\left(\frac 13,\frac 23\right)$. Is there any easier solution?
 [/INST] For the second integral, make the substitution $ \displaystyle x = \frac{u}{1+u}$.
Then
$$ \begin{align} 2 \int_{0}^{1} \frac{dx}{(2-x) \sqrt[3]{x^2(1-x)}} &= 2 \int_{0}^{\infty}  \frac{du}{(2+u)u^{2/3}} \\ &= \int_{0}^{\infty}  \frac{du}{\left(1+ \frac{u}{2}\right)u^{2/3}} \\ &= \frac{2}{2^{2/3}} \int_{0}^{\infty} \frac{1}{(1+t)t^{2/3}} \, dt \\ &= \frac{2}{2^{2/3}} \int_{0}^{\infty} \frac{t^{1/3-1}}{(1+t)^{1/3+2/3}} \, dt \\ &= \frac{2}{2^{2/3}} B \left(\frac{1}{3}, \frac{2}{3} \right) \tag{1} \\&= \frac{2}{2^{2/3}} \frac{\Gamma \left(\frac{1}{3} \right) \Gamma \left(1- \frac{1}{3} \right)}{\Gamma(1)} \\ &=  \frac{2}{2^{2/3}} \frac{2 \pi}{\sqrt{3}} \tag{2} \\ &= \frac{2^{4/3} \pi}{\sqrt{3}}. \end{align}$$
$(1)$ https://en.wikipedia.org/wiki/Beta_function#Properties
$(2)$ https://en.wikipedia.org/wiki/Gamma_function#General
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