Contour Integration $\int_0^1\frac1{\sqrt[n]{1-x^n}}dx$ I want to compute:
$$\int^{1}_{0}\frac{1}{\sqrt[n]{1-x^n}}dx$$
for natural $n>1$ using Residue Calculus.
I am thinking of using some kind of a keyhole or bone contour that could go around the $n$th roots of unity (singularities in this case). The problem is I believe it is not clear how to define a suitable branch (or branches) of $\log$ in this region for it to work, also considering we only care about the segment from $0$ to $1$.

In questions like this one, in order to avoid the problems of defining the right branch of the logarithm or the $n$th root, I suggest to, first start with a change of variables and to use Residue Theorem afterwards.
So, here how I do this. First the change of variables $x^n=\dfrac{e^t}{1+e^t}$ we get
$$
I_n~{\buildrel {\rm def}\over =}~\int_0^1\frac{dx}{\root{n}\of{1-x^n}}=\frac{1}{n}\int_{-\infty}^\infty \frac{e^{t/n}}{1+e^t}dt
$$
Next we integrate $F(z)=\dfrac{e^{z/n}}{n(1+e^z)}$ on the rectangle $\Gamma_R$ with vertexes $-R$, $R$,$R+2i\pi$ and $-R+2i\pi$. Letting $R$ tend to $+\infty$ we get
$$
I_n-e^{2i\pi/n}I_n=2i\pi~\hbox{Res}(F(z),i\pi)=-2i\frac{\pi}{n} e^{i\pi/n}.
$$
This yields 
$$I_n=\frac{\pi}{n\sin(\pi/n)}.$$
Note that we didn't use the fact that $n$ is a natural number. This is valid for any real $n>1$.
