= Solution
Use the <branch of the complex logarithm> with $0<\arg z<2\pi$, so the <branch cut> is the positive real axis, and take $\sqrt z=\exp(\operatorname{Log}z/2)$ on the same slit plane. Apply the <residue theorem> to
$$
F(z)=\frac{\operatorname{Log}z}{(z^2+1)\sqrt z}
$$
on a positively oriented <keyhole contour> with inner radius $\rho<1$ and outer radius $R>1$. The upper bank travels from $\rho$ to $R$, the large circle counterclockwise, the lower bank from $R$ to $\rho$, and the small circle clockwise. One may first keep the banks at small nonzero angles and then take their limits.
On the upper bank, $\operatorname{Log}x=\log x$ and $\sqrt x$ is positive. On the lower bank, the corresponding limits are $\log x+2\pi i$ and $-\sqrt x$. Reversing the lower-bank orientation therefore makes the sum of the bank <integrals> tend to
$$
2I+2\pi iJ,\qquad I=\int_0^\infty\frac{\log x}{(x^2+1)\sqrt x}\,dx,\qquad J=\int_0^\infty\frac{dx}{(x^2+1)\sqrt x}.
$$
Both are absolutely convergent: near zero compare with $x^{-1/2}(1+|\log x|)$, and near infinity with $x^{-5/2}(1+\log x)$. On the large circle, $|z^2+1|\geq R^2-1$, so its contribution is $O((\log R+1)R^{-3/2})$. On the small circle it is $O(\sqrt\rho(1+|\log\rho|))$. Both tend to zero.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2012-keyhole.png]
{title=The <keyhole contour> and the two boundary values on its <branch cut>}
The two simple <poles> are $i$ and $-i$. On the chosen branch their arguments are $\pi/2$ and $3\pi/2$, respectively. Consequently their <residues> are
$$
\operatorname{Res}(F,i)=\frac{\pi}{4\sqrt2}(1-i),\qquad \operatorname{Res}(F,-i)=\frac{3\pi}{4\sqrt2}(1+i).
$$
For example, the first follows from $\operatorname{Log}i=i\pi/2$, $\sqrt i=e^{i\pi/4}$ and the <derivative> $2i$ of the denominator factor $z^2+1$. The second uses $\sqrt{-i}=e^{3\pi i/4}$, rather than the value from the principal square-root branch. Adding and applying the <residue theorem> yields
$$
2I+2\pi iJ=2\pi i\left(\frac\pi{\sqrt2}+\frac{i\pi}{2\sqrt2}\right)=-\frac{\pi^2}{\sqrt2}+i\sqrt2\pi^2.
$$
As $J$ is real, taking real parts gives
$$
\boxed{I=-\frac{\pi^2}{2\sqrt2}.}
$$
The imaginary part also gives $J=\pi/\sqrt2$, an independent consistency check on the orientation and branch factors.
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