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$\ds{}$
\begin{align}
I&\equiv\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x}
=\int_{\infty}^{0}{\ln\pars{1/x} \over 1/x^{2} + 1}\,\pars{-\,{\dd x \over x^{2}}}
=-\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x}
=-I\ \imp 2I = 0
\\[3mm]&\imp I = 0\quad\imp\quad
\color{#66f}{\large\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x = 0}
\end{align}