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\begin{align}
\int_{0}^{x}{\arctan\pars{t} \over t}\,\dd t&=\ln\pars{x}\arctan\pars{x}
-\int_{0}^{x}{\ln\pars{t} \over t^{2} + 1}\,\dd t
\\[3mm]&=\ln\pars{x}\arctan\pars{x}
-\lim_{\mu\to 0}\partiald{}{\mu}\int_{0}^{x}{t^{\mu} \over t^{2} + 1}\,\dd t
\\[3mm]&=\ln\pars{x}\arctan\pars{x}
-\lim_{\mu\to 0}\partiald{}{\mu}\int_{0}^{x}{t^{\mu/2} \over t + 1}\,\half\,t^{-1/2}\,\dd t
\\[3mm]&=\ln\pars{x}\arctan\pars{x}
-
\half\,\lim_{\mu\to 0}\partiald{}{\mu}
\int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t\tag{1}
\end{align}
With
$\ds{\xi \equiv {1 \over t + 1}\quad\iff\quad t = {1 \over \xi} - 1}
= {1 - \xi \over \xi}$
\begin{align}
\int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t&=
\int_{1}^{1/\pars{x + 1}}\xi\pars{1 - \xi \over \xi}^{\pars{\mu - 1}/2}\,
\pars{-\,{\dd \xi \over \xi^{2}}}
\\[3mm]&=
\int^{1}_{1/\pars{x + 1}}\xi^{-\pars{\mu + 1}/2}\pars{1 - \xi}^{\pars{\mu - 1}/2}\,
\dd \xi\\[3mm]&={\rm B}\pars{{1 - \mu \over 2},{1 + \mu \over 2}}
-{\rm B}_{1/\pars{x + 1}}\pars{{1 - \mu \over 2},{1 + \mu \over 2}}\tag{2}
\end{align}
where ${\rm B}\pars{p,q}$ and ${\rm B}_{x}\pars{p,q}$ are the Beta and the
Incomplete Beta functions, respectively. ${\rm B}\pars{p,q}$ satisfies
$\ds{{\rm B}\pars{p,q} = {\Gamma\pars{p}\Gamma\pars{q} \over \Gamma\pars{p +} q}}$
such that
$$
{\rm B}\pars{{1 - \mu \over 2},{1 + \mu \over 2}}
={\Gamma\pars{1/2 - \mu/2}\Gamma\pars{1/2 + \mu/2} \over \Gamma\pars{1}}
={\pi \over \sin\pars{\pi\bracks{1/2 + \mu/2}}}
={\pi \over \cos\pars{\pi\mu/2}}
$$
$\pars{2}$ is reduced to:
$$
\int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t
=\pi\sec\pars{\pi\mu \over 2}-{\rm B}_{1/\pars{x + 1}}\pars{{1 - \mu \over 2},{1 + \mu \over 2}}
$$
In terms of the Hipegeometric Function:
$$
\int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t
=\pi\sec\pars{\pi\mu \over 2}
-
2\,{\pars{1 + x}^{\pars{\mu - 1}/2} \over 1 - \mu}\
_{2}{\rm F}_{1}\pars{{1 - \mu \over 2},{1 - \mu \over 2};{3 - \mu \over 2};
{1 \over x + 1}}
$$
and
\begin{align}
&\lim_{\mu \to 0}\partiald{}{\mu}
\int_{0}^{x}{t^{\pars{\mu - 1}/2} \over t + 1}\,\dd t
\\[3mm]&=-\,\bracks{\ln\pars{1 + x} + 2}\arcsin\pars{1 \over \root{1 + x}}
\\[3mm]&\phantom{=}-
\\[3mm]&\phantom{=}
{2 \over \root{1 + x}}\lim_{\mu \to 0}\partiald{}{\mu}\
_{2}{\rm F}_{1}\pars{{1 - \mu \over 2},{1 - \mu \over 2};{3 - \mu \over 2};
{1 \over x + 1}}
\end{align}
The answer is found by replacing this expression in $\pars{1}$.