$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 $\pars{1}$.