Integral $\int_0^1\sqrt[2\,n\,]{\frac x{1-x}}\,\mathrm dx$ I'm trying to express this integral
$$
\int_0^1\sqrt[2\,n\,]{\frac x{1-x}}\,\mathrm dx
$$
for any $n\in\mathbb N$. I've tried integral by substitution and partial fraction decomposition but it does not seem to lead to the solution.
Do you have any advice ?
My point is to calculate this integral:

For this I divided it in two integrals:

I used the substitution method twice. Is the result right now ?
Was there an easier way to express this integral ?
Thanks a lot !
 A: This is what is known as a beta function.  Generally:
$$\int_0^1 dx \, x^{a-1} (1-x)^{b-1} = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} $$
In your case,
$$a=1+\frac1{2 n}$$
$$b = 1-\frac1{2 n} $$
So the result is
$$\Gamma \left ( 1+\frac1{2 n} \right ) \Gamma \left ( 1-\frac1{2 n} \right ) $$
which may be simplified to
$$-\frac1{2 n} \Gamma \left ( 1+\frac1{2 n} \right ) \Gamma \left ( -\frac1{2 n} \right ) = \frac{\pi}{2 n \sin{\frac{\pi}{2 n}}}$$
A: An alternative method arises by substituting $y=x/(1-x)$; then the integral becomes
$$\int_0^{\infty} dy \, \frac{y^{1/(2 n)} }{(1+y)^2} $$
This may be evaluated by contour integration using the residue theorem.  Consider the contour integral
$$\oint_C dz \, \frac{z^{1/(2 n)} }{(1+z)^2} $$
where $C$ is a keyhole contour about the positive real axis.  The contour integral may be shown to be equal to
$$\left (1-e^{i \pi/n}\right ) \int_0^{\infty} dx \, \frac{x^{1/(2 n)} }{(1+x)^2} $$
By the residue theorem, the contour integral is also equal to
$$-i 2 \pi \frac1{2 n} e^{i \pi/(2 n)} $$
The integral is thus equal to
$$\int_0^{\infty} dx \, \frac{x^{1/(2 n)} }{(1+x)^2} = \frac{\pi}{2 n \sin{\frac{\pi}{2 n}}} $$
A: My point was to calculate this integral:

For this I divided it in two integrals:

I used the substitution method twice. Is the result right now ?
Was there an easier way to express this integral ?
[I don't know if it's the right place to ask an other question... I'm sorry if it's not]
