= Solution
Use the same indented upper-semicircle contour and <branch of the complex logarithm>, now for $F(z)=z^{1/2}\operatorname{Log}z/(1+z^2)$. Write $J=\int_0^\infty x^{1/2}\log x/(1+x^2)\,dx$. On the negative axis, $z^{1/2}=i|z|^{1/2}$ and $\operatorname{Log}z=\log|z|+i\pi$. Consequently the two straight contributions tend to
$$
(1+i)J-\pi I(1/2).
$$
The outer arc is $O(R^{-1/2}\log R)$ and the small arc is $O(\varepsilon^{3/2}(|\log\varepsilon|+1))$, so neither contributes in the limit. At the enclosed <pole> $i$,
$$
\operatorname{Res}_{z=i}F=\frac{e^{i\pi/4}(i\pi/2)}{2i}=\frac\pi4e^{i\pi/4}.
$$
The <residue theorem> gives $(1+i)J-\pi I(1/2)=i\pi^2e^{i\pi/4}/2$. Part (a) gives $I(1/2)=\pi/\sqrt2$. Substituting and using $e^{i\pi/4}=(1+i)/\sqrt2$, the right side after moving the $I$ term is $\pi^2(1+i)/(2\sqrt2)$. Hence
$$
\boxed{J=\frac{\pi^2}{2\sqrt2}.}
$$
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