TITLE: Find $\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$
QUESTION [5 upvotes]: Need solve the next integral
$$\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$$
Tried something with Laurent’s series, but i can’t conclude anything.
Thanks

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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Lets consider
\begin{equation}
\int_{\mathcal{C}}{z^{\mu} \over z - a}\,\dd z\,,\qquad \mu \in \pars{-1,0}\,,\quad a > 0\label{1}\tag{1}
\end{equation}
$\ds{\mathcal{C}}$ is a key-hole contour where $\ds{z^{\mu}}$ is its principal branch.
Then,
\begin{align}
2\pi\ic a^{\mu} & =
-\int_{-\infty}^{0}{\pars{-x}^{\mu}\expo{\ic\pi\mu} \over x - a}\,\dd x -
\int_{0}^{-\infty}{\pars{-x}^{\mu}\expo{-\ic\pi\mu} \over x - a}\,\dd x
\\[5mm] & =
\expo{\ic\pi\mu}\int_{0}^{\infty}{x^{\mu} \over x + a}\,\dd x -
\expo{-\ic\pi\mu}\int_{0}^{\infty}{x^{\mu} \over x + a}\,\dd x
\\[5mm] & =
2\ic\sin\pars{\pi\mu}\int_{0}^{\infty}{x^{\mu} \over x + a}\,\dd x
\\[5mm] \implies &
\bbx{\int_{0}^{\infty}{x^{\mu} \over x + a }\,\dd x = {\pi a^{\mu} \over \sin\pars{\pi\mu}}}\label{2}\tag{2}
\end{align}

Derive (\ref{2}) respect of $\ds{a}$:
$\ds{\int_{0}^{\infty}{x^{\mu} \over \pars{x + a}^{\, 2}}\,\dd x = 
-\pi\,{\mu a^{\mu - 1} \over \sin\pars{\pi\mu}}}$

Set $\ds{a = 1}$:
$\ds{\int_{0}^{\infty}{x^{\mu} \over \pars{x + 1}^{\, 2}}\,\dd x = 
-\,{\pi\mu \over \sin\pars{\pi\mu}}}$

Derive respect of $\ds{\mu}$:
$\ds{\int_{0}^{\infty}{x^{\mu}\ln\pars{x} \over \pars{x + a}^{\, 2}}\,\dd x = 
-\pi\bracks{1 - \pi\mu\cot\pars{\pi\mu}\csc\pars{\pi\mu}}}$

Takes the limit $\ds{\mu \to -1/2}$:
$$
\bbox[15px,#ffd,border:1px solid navy]{
\int_{0}^{\infty}{\ln\pars{x} \over
\root{x}\pars{x + a}^{\, 2}}\,\dd x = {\large -\pi}}\\
$$