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$\ds{\int_{0}^{\infty}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x
={\rm G} + {1 \over 4}\,\pi\,\ln\pars{2}\quad\mbox{where}\quad
{\rm G} \equiv \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{2}}
= 0.9159\ldots}$ is the Catalan Constant.
Let's consider the integral
$$
{\cal I} \equiv \int_{\rm C}{\ln^{2}\pars{z} \over z^{2} + 2z + 2}\,\dd z
\quad\mbox{where}\quad
\left\vert%
\begin{array}{l}
{\rm C}\ \mbox{is a 'key-hole' contour which takes care of the}\
\\
{\large \ln}-\mbox{branch cut}.
\end{array}\right.
$$
Zeros of $\ds{z^{2} + 2z + 2=0}$ are given by
$\ds{z_{\pm} = -1 \pm \ic = \root{2}\expo{\pm 3\ic\pi/4}}$.
\begin{align}
{\cal I}&=2\pi\ic\,\bracks{{\ln^{2}\pars{z_{+}} \over z_{+} - z_{-}}
+ {\ln^{2}\pars{z_{-}} \over z_{-} - z_{+}}}
=2\pi\ic\,{%
\bracks{\ln\pars{2}/2 + 3\ic\pi/4}^{2}
-\bracks{\ln\pars{2}/2 - 3\ic\pi/4}^{2}\over 2\ic}
\\[3mm]&=\pi\bracks{\ic\,{3\pi \over 2}\,\ln\pars{2}}
={3 \over 2}\,\pi^{2}\ln\pars{2}\,\ic\tag{1}
\end{align}
\begin{align}
{\cal I}&=
\int_{-\infty}^{0}{\bracks{\ln^{2}\pars{-x} + \ic\pi} \over x^{2} + 2x + 2}\,\dd x
+
\int_{0}^{-\infty}{\bracks{\ln^{2}\pars{-x} - \ic\pi} \over x^{2} + 2x + 2}\,\dd x
=-4\pi\ic\int_{0}^{-\infty}{\ln\pars{-x} \over x^{2} + 2x + 2}\,\dd x
\\[3mm]&=4\pi\ic\int_{0}^{\infty}{\ln\pars{x} \over x^{2} - 2x + 2}\,\dd x
=4\pi\ic\int_{0}^{1}{\ln\pars{x} \over x^{2} - 2x + 2}\,\dd x
+
4\pi\ic\int_{1}^{\infty}{\ln\pars{x} \over x^{2} - 2x + 2}\,\dd x
\\[3mm]&=4\pi\ic\int_{0}^{1}{\ln\pars{x} \over x^{2} - 2x + 2}\,\dd x
+
4\pi\ic\int_{0}^{\infty}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x
\end{align}
$$
\!\!\!\!\!\!\!\!\!\!\int_{0}^{\infty}{\ln\pars{1 + x} \over x^{2} + 1}\,\dd x
={{\cal I} \over 4\pi\ic} - \int_{0}^{1}{\ln\pars{x} \over x^{2} - 2x + 2}\,\dd x
={3 \over 8}\,\pi\ln\pars{2}
-\color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} - 2x + 2}\,\dd x}\tag{2}
$$
where we used result $\pars{1}$.
$\ds{z_{\pm} = 1 \pm \ic}$ are the zeros of
$\ds{z^{2} - 2z + 2 = 0}$.
$\ds{\color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} - 2x + 2}\,\dd x}}$ is evaluated as follows:
\begin{align}
&\color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} - 2x + 2}\,\dd x}=
\int_{0}^{1}\ln\pars{x}\pars{{1 \over x - z_{+}} - {1 \over x - z_{-}}}
\,{1 \over z_{+} - z_{-}}\,\dd x
=\Im\int_{0}^{1}{\ln\pars{x} \over x - z_{+}}\,\dd x
\\[3mm]&=-\Im\int_{0}^{1/z_{+}}{\ln\pars{z_{+}x} \over 1 - x}\,\dd x
=-\Im\pars{\ln\pars{z_{+}}\int_{0}^{1/z_{+}}{\dd x \over 1 - x}
+\int_{0}^{1/z_{+}}{\ln\pars{x} \over 1 - x}\,\dd x}
\\[3mm]&=-\Im\pars{-\ln\pars{z_{+}}\ln\pars{1 - {1 \over z_{+}}}
-\ln\pars{1 \over z_{+}}\ln\pars{1 - {1 \over z_{+}}}
+\int_{0}^{1/z_{+}}{\ln\pars{1 - x} \over x}\,\dd x}
\\[3mm]&=\Im\int_{0}^{1/z_{+}}{{\rm Li}_{1}\pars{x} \over x}\,\dd x
=\Im{\rm Li}_{2}\pars{1 \over z_{+}} = \Im{\rm Li}_{2}\pars{\half - \half\,\ic}
=\color{#00f}{-{\rm G} + {1 \over 8}\,\pi\ln\pars{2}}\qquad\pars{3}
\end{align}
where $\ds{{\rm Li}_{s}\pars{z}}$ is the Polylogarithm Function and
$\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$. We used the recursive relation
$\ds{{\rm Li}_{s + 1}\pars{z} = \int_{0}^{z}{{\rm Li}_{s}\pars{x} \over x}\,\dd x}$.
The $\ds{\color{#00f}{\mbox{last result}}}$ is found with the series representation of $\ds{{\rm Li}_{s}\pars{z}}$.
By replacing $\pars{3}$ in $\pars{2}$, we found:
$$\color{#00f}{\large%
\int_{0}^{\infty}{\ln\pars{1 + x} \over x^{2} + 1}\,\dd x
=G + {1 \over 4}\,\pi\ln\pars{2}}
$$
