= Solution
The finite singularities are square-root <Branch point>[branch points] at $z=\pm1$. Choose the cut between them and the branch $f(z)\sim z$ at infinity. The two banks then have values $+i\sqrt{1-x^2}$ above and $-i\sqrt{1-x^2}$ below. A counterclockwise loop around the cut traverses its upper bank right-to-left, so its <integral> is $-2iI$, where $I$ is the required real <integral>. At large <radius>,
$$
f(z)=z-\frac1{2z}+O(z^{-3}),\qquad\oint f(z)dz=-\pi i.
$$
Therefore $\boxed{I=\pi/2}$. Infinity is not another square-root <branch point>: this branch has an ordinary <pole> there on its sheet. Small detours about the finite endpoints vanish as their <radius> to power $3/2$.
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