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\begin{align}
&\color{#f00}{\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\bracks{x + y}\verts{1 - xy}} \over \pars{1 + x^{2}}\pars{1 +y^{2}}}
\,\dd x\,\dd y}\ =\ \overbrace{\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y}^{\ds{=\ 0}}
\\[5mm] +&\
\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{1 + y/x} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y +
\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\verts{1 - xy}} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y
\end{align}
In the second integral, in the RHS, lets $\ds{x\ \mapsto\ 1/x}$:
\begin{align}
&\color{#f00}{\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\bracks{x + y}\verts{1 - xy}} \over \pars{1 + x^{2}}\pars{1 +y^{2}}}
\,\dd x\,\dd y}
\\[5mm] = &\
\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{1 + y/x} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y +
\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\verts{1 - y/x}} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y
\\[5mm] = &\
\int_{0}^{\infty}{1 \over x^{2} + 1}\int_{0}^{\infty}
{\ln\pars{\verts{1 - y^{2}/x^{2}}} \over 1 + y^{2}}\,\dd y\,\dd x
\,\,\,\stackrel{y/x\ \mapsto\ y}{=}\,\,\,
\int_{0}^{\infty}{x \over x^{2} + 1}\int_{0}^{\infty}
{\ln\pars{\verts{1 - y^{2}}} \over 1 + x^{2}y^{2}}\,\dd y\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}\ln\pars{\verts{1 - y^{2}}}\ \overbrace{\int_{0}^{\infty}
{x \over \pars{x^{2} + 1}\pars{y^{2}x^{2} + 1}}\,\dd x}
^{\ds{\ln\pars{y} \over y^{2} - 1}}\ \,\dd y\ =\
\int_{0}^{\infty}{\ln\pars{\verts{1 - y^{2}}}\ln\pars{y} \over y^{2} - 1}\,\dd y
\end{align}
Now, we split the integral along $\ds{\pars{0,1}}$ and $\ds{\pars{1,\infty}}$.
Later on, we make the substitution $\ds{y \mapsto 1/y}$ in the second integral
$\pars{~\mbox{along}\ \pars{1,\infty}~}$:
\begin{align}
&\color{#f00}{\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\bracks{x + y}\verts{1 - xy}} \over \pars{1 + x^{2}}\pars{1 +y^{2}}}
\,\dd x\,\dd y}
\\[5mm] = &\
\int_{0}^{1}{\ln\pars{1 - y^{2}}\ln\pars{y} \over y^{2} - 1}\,\dd y +
\int_{0}^{1}{\bracks{\ln\pars{1 - y^{2}} - 2\ln\pars{y}}\ln\pars{y} \over
y^{2} - 1}\,\dd y
\\[5mm] = &\
2\int_{0}^{1}
{\ln\pars{1 - y^{2}}\ln\pars{y} - \ln^{2}\pars{y} \over y^{2} - 1}\,\dd y
\,\,\,\stackrel{y^{2}\ \mapsto\ y}{=}\,\,\,
2\int_{0}^{1}
{\ln^{2}\pars{y}/4 - \ln\pars{1 - y}\ln\pars{y}/2 \over 1 - y}\,{1 \over 2}\,y^{-1/2}\dd y
\\[5mm] = &\
{1 \over 4}\ \underbrace{%
\int_{0}^{1}{y^{-1/2}\ln^{2}\pars{y} \over 1 - y}\,\dd y}
_{\ds{=\ 14\zeta\pars{3}}}\ -\
{1 \over 2}\ \underbrace{\int_{0}^{1}{y^{-1/2}\ln\pars{y}\ln\pars{1 - y} \over 1 - y}\,\dd y}_{\ds{=\ -\pi^{2}\ln\pars{2} + 7\zeta\pars{3}}}\ =\
\color{#f00}{{1 \over 2}\,\pi^{2}\ln\pars{2}}
\end{align}
The last two integrals can be straightforward reduced to derivatives of the Beta Function and evaluated at suitable limits.
