 Solution

The finite singularities are square-root branch points at $z=\pm 1$. 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,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 $3/2$.