= Solution
Use the <branch cut> $[-1,1]$ and the <branch of a multivalued function> normalized by $\sqrt{z^2-1}\sim z$. The <Laurent series> from the preceding solution has $z^{-1}$ coefficient $3/8$. Consequently every sufficiently large positively oriented circle gives
$$
\boxed{\oint_{C_\infty}f(z)\,dz=\frac{3\pi i}{4}.}
$$
Although $f$ grows cubically, its other <Laurent series> terms have zero <contour integral>. Choosing the opposite <branch of a multivalued function> reverses the sign; neither the <branch of a multivalued function> nor the <orientation> can be omitted from the answer.
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