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 !

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
$$\int_0^{\infty} dx \, \frac{x^{1/(2 n)} }{(1+x)^2} = \frac{\pi}{2 n \sin{\frac{\pi}{2 n}}} $$
