= Solution
A <branch point> is a point around which <analytic continuation> changes the value of a multivalued function. For $z^{1/2}=r^{1/2}e^{i\theta/2}$, one circuit around zero changes the sign. A <branch cut> removes paths producing this ambiguity; for example $-\pi<\arg z<\pi$ gives a single holomorphic <square root> off the nonpositive <real axis>. The <position> of a cut is a convention, while the <branch point> is intrinsic.
Set $q(z)=(z-1)/(z+1)$. For $z$ outside the closed segment joining $-1$ to $1$, $q(z)$ avoids the nonpositive <real axis>. Therefore the <principal square root> of $q$ defines a <holomorphic function> there, with the <normalization> $f(\infty)=1$. The endpoints are <branch points>. Expanding at infinity,
$$
f(z)=\left(1-\frac2{z+1}\right)^{1/2}=1-\frac1z+O(z^{-2}).
$$
For a positively oriented contour surrounding the cut, its <integral> equals the <integral> on a sufficiently large positively oriented circle, since the intervening annulus is holomorphic. Thus
$$
\boxed{I_\Gamma=-2\pi i.}
$$
On the upper and lower banks, $f_+(x)=+i\sqrt{(1-x)/(1+x)}$ and $f_-(x)=-i\sqrt{(1-x)/(1+x)}$. The upper bank is traversed from $1$ to $-1$, and the lower bank from $-1$ to $1$. Their sum is $-2i$ <times> the required real <integral>. Small endpoint arcs vanish because their singularities have exponent less than one. Consequently
$$
\boxed{\int_{-1}^{1}\sqrt{\frac{1-x}{1+x}}\,dx=\pi.}
$$
For $\alpha=1/n$, use $f_\alpha=q^\alpha$ with the same <normalization>. Its expansion is $1-2\alpha/z+O(z^{-2})$, and its bank values are $e^{\pm i\pi\alpha}[(1-x)/(1+x)]^\alpha$. Equating the <contour integral> $-4\pi i\alpha$ with $-2i\sin(\pi\alpha)$ <times> the real <integral> gives
$$
\boxed{\int_{-1}^{1}\left(\frac{1-x}{1+x}\right)^{1/n}dx
=\frac{2\pi}{n\sin(\pi/n)}.}
$$
The condition $n\ge2$ ensures endpoint integrability. Choosing the other square-root branch reverses $I_\Gamma$, but also reverses both bank values, leaving the real <integral> unchanged.
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