Evaluating $\int_{0}^\infty \frac{\log x \, dx}{\sqrt x(x^2+a^2)^2}$ using contour integration I need your help with this integral:
$$\int_{0}^\infty \frac{\log x \, dx}{\sqrt x(x^2+a^2)^2}$$
where $a>0$. I have tried some complex integration methods, but none seems adequate for this particular one.
Is there a specific method for this kind of integrals? What contour should I use?

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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\color{#f00}{%
\int_{0}^{\infty}{\ln\pars{x} \over \root{x}\pars{x^{2} + a^{2}}^{2}}\,\dd x}
\,\,\,\stackrel{x\ \to\ x^{1/2}}{=}\,\,\,
{1 \over 4}\int_{0}^{\infty}
{x^{-3/4}\,\ln\pars{x} \over \pars{x + a^{2}}^{2}}\,\dd x
\\[5mm] = &\
-\,{1 \over 4}\,\partiald{}{\pars{a^{2}}}\int_{0}^{\infty}
{x^{-3/4}\,\ln\pars{x} \over x + a^{2}}\,\dd =
-\,{1 \over 8\verts{a}}\,\partiald{}{\verts{a}}\int_{0}^{\infty}
{x^{-3/4}\,\ln\pars{x} \over x + a^{2}}\,\dd x
\\[5mm] = &\
-\,{1 \over 8\verts{a}}\,\partiald{}{\verts{a}}\bracks{%
\lim_{\mu \to -3/4}\,\,\partiald{}{\mu}
\int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x}\tag{1}
\end{align}

With the branch-cut $\ds{z^{\mu} = \verts{z}^{\mu}\exp\pars{\ic\,\mathrm{arg}\pars{z}\mu}\,,\ 0 < \mathrm{arg}\pars{z} < 2\pi\,,\ z \not = 0}$, the integral is performed along a key-hole contour. Namely,
\begin{align}
2\pi\ic\,\verts{a}^{2\mu}\exp\pars{\ic\pi\mu} & =
\int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x +
\int_{\infty}^{0}{x^{\mu}\exp\pars{2\pi\mu\ic} \over x + a^{2}}\,\dd x
\\[3mm] & =
-\exp\pars{\ic\pi\mu}\bracks{2\ic\sin\pars{\pi\mu}}
\int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x
\\[5mm] 
\imp\ \int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x & =
-\pi\,\verts{a}^{2\mu}\csc\pars{\pi\mu}
\end{align}

Plug this result in $\pars{1}$:
$$
\color{#f00}{%
\int_{0}^{\infty}{\ln\pars{x} \over \root{x}\pars{x^{2} + a^{2}}^{2}}\,\dd x} =
\color{#f00}{{\root{2} \over 16}\,\pi\,{%
6\ln\pars{\verts{a}} - 3\pi - 4 \over \verts{a}^{7/2}}}
$$
