= Solution
Use a counterclockwise <keyhole contour> around the positive real <branch cut>, with small radius $\varepsilon$ and large radius $R$. On the upper bank $\arg z=0$, so the integral tends to $I$. On the lower bank $\arg z=2\pi$, $z^{1/2}=-\sqrt{x}$, giving the negative of the upper-bank integrand. That bank is traversed from $R$ to $\varepsilon$, so its contribution also tends to $I$.
The outer circle contributes $O(R^{-3/2})$, since $f=O(R^{-5/2})$ and its length is $O(R)$. The small circle contributes $O(\varepsilon^{1/2})$, since $f=O(\varepsilon^{-1/2})$. Both vanish. Hence
$$
\boxed{C=\lim_{\varepsilon\to0,\,R\to\infty}\oint_{\Gamma_{\varepsilon,R}}f(z)\,dz=2I.}
$$
The contour encloses both double <poles> $\pm i$ but excludes the branch point.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2-q4-keyhole.png]
{title=Keyhole contour around the positive real branch cut}
{description=The upper bank runs outwards and the lower bank runs inwards. The large circle is counterclockwise and the small circle is clockwise; both <poles> lie inside the contour. The gaps are enlarged for visibility.}
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