We integrate with $a$ a positive real
$$f(z) = \frac{\mathrm{Log}(z)}{(z-ai)^2 (z+ai)^2}
\exp(-1/2\times \mathrm{Log}(z))$$
around a keyhole contour with the slot on the positive real axis,
which is also where the branch cut of the logarithm is located
(argument of the logarithm is between $0$ and $2\pi$). Now for the
large circle we get $\lim_{R\to\infty} 2\pi R \log R / \sqrt{R} /R^4 =
0$ so there is no contribution in the limit. For the small circle
around the origin we find $\lim_{\epsilon\to\ 0} 2\pi \epsilon \log
\epsilon / \sqrt\epsilon /a^4 = 0$ so there is no contribution here
either.
We get for the upper line segment
$$\int_0^\infty \frac{\log x}{(x^2+a^2)^2}
\exp(-1/2\times \log x) \; dx$$
which is our target integral, call it $J$. The lower line segment
contributes
$$-\int_0^\infty \frac{\log x + 2\pi i}{(x^2+a^2)^2}
\exp(-1/2\times \log x) \exp(-1/2\times 2\pi i) \; dx
\\ = \int_0^\infty \frac{\log x + 2\pi i}{(x^2+a^2)^2}
\exp(-1/2\times \log x) \; dx
\\ = J + 2\pi i \int_0^\infty \frac{1}{(x^2+a^2)^2}
\exp(-1/2\times \log x) \; dx = J + 2\pi i K$$
where $J$ and $K$ are real numbers. We thus have
$$2J + 2\pi i K = 2\pi i \mathrm{Res}_{z=ai} f(z)
+ 2\pi i \mathrm{Res}_{z=-ai} f(z)$$
or
$$J + \pi i K = \pi i \mathrm{Res}_{z=ai} f(z)
+ \pi i \mathrm{Res}_{z=-ai} f(z)$$
With this setup we do not actually need to compute $K$ as it must
correspond to the imaginary part of the contribution from the two
residues. We get for the first residue at $z=ai$ the derivative
$$\frac{1}{z} \frac{1}{(z+ai)^2} \exp(-1/2\times \mathrm{Log}(z))
-2 \mathrm{Log}(z)
\frac{1}{(z+ai)^3} \exp(-1/2\times \mathrm{Log}(z))
\\ + \mathrm{Log}(z) \frac{1}{(z+ai)^2}
\exp(-1/2\times \mathrm{Log}(z)) \times -\frac{1}{2} \frac{1}{z}.$$
With the branch of the logarithm we find $\mathrm{Log}(ai) =
\log a + \pi i/2$, getting
$$\frac{1}{\sqrt{a}} \exp(-\pi i/4)
\\ \times \left(\frac{1}{ai} \times - \frac{1}{4 a^2}
+ (2\log a + \pi i) \frac{1}{8 i a^3}
+ (\log a + \pi i/2) \times - \frac{1}{4 a^2}
\times -\frac{1}{2 ai}\right)
\\ = \frac{1}{\sqrt{a}} \exp(-\pi i/4) \frac{1}{8i a^3}
(3\log a + 3\pi i/2 - 2).$$
We also have $\mathrm{Log}(-ai) = \log a + 3 \pi i/2$, getting
for the second residue at $z=-ai$
$$\frac{1}{\sqrt{a}} \exp(-3\pi i/4)
\\ \times \left(- \frac{1}{ai} \times - \frac{1}{4 a^2}
- (2\log a + 3\pi i) \frac{1}{8 i a^3}
+ (\log a + 3\pi i/2) \times - \frac{1}{4 a^2}
\times \frac{1}{2 ai}\right)
\\ = \frac{1}{\sqrt{a}} \exp(-3\pi i/4) \frac{1}{8i a^3}
(- 3\log a - 9\pi i/2 + 2).$$
With $\exp(-\pi i/4) =
\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$
and $\exp(-3\pi i/4) =
-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$
and factoring out $\frac{1}{\sqrt{a}} \frac{1}{8i a^3}$
we get three contributions, which are
$$\sqrt{2} (3\log a - 2) + 3\pi i (\sqrt{2} + i\sqrt{2}/2)$$
Combine these and multiply by $\pi i$ to get
$$\frac{\pi}{8\sqrt{a} a^3}
(\sqrt{2} (3\log a - 2) + 3\pi i (\sqrt{2} + i\sqrt{2}/2)).$$
We extract the real part as promised and obtain
$$\frac{\sqrt{2}\pi}{8\sqrt{a} a^3}
(3\log a - 2 - 3\pi/2)$$
or
$$\bbox[5px,border:2px solid #00A000]{
\frac{\pi}{4\sqrt{2} a^{7/2}}
\left(3\log a - 2 - \frac{3}{2}\pi\right).}$$
matching the result by @Jack D'Aurizio.
As a bonus we have shown that
$$\int_0^\infty \frac{1}{\sqrt{x} (x^2+a^2)^2} dx
= \frac{1}{\pi} \frac{\pi}{8\sqrt{a} a^3} 3\sqrt{2}\pi
= \frac{3\pi}{4\sqrt{2} a^{7/2}}.$$