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$\ds{\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}}
\,{\dd x \\pars{1 + x^{2}}\ln\pars{x}}:\ {\Large ?}}$.
\begin{equation}
\mbox{Note that}\quad
\begin{array}{|l|}\hline\mbox{}\\
\ds{\quad\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}}
\,{\dd x \over \pars{1 + x^{2}}\ln\pars{x}} =\quad}
\\[3mm]
\ds{\quad\int_{0}^{\infty}
{\ln\pars{1 + x^{11}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x -
\int_{0}^{\infty}
{\ln\pars{1 + x^{3}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x\quad}
\\ \mbox{}\\ \hline
\end{array}
\label{1}\etiqueta{1}
\end{equation}
Con $\ds{\mu > 0}$:
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}
{\ln\pars{1 + x^{\mu}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x}}
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\int_{\infty}^{0}
{\ln\pars{1 + 1/x^{\mu}} \over \pars{1 + 1/x^{2}}\ln\pars{1/x}}
\pars{-\,{\dd x \over x^{2}}}
\\[5mm] = &\
-\int_{0}^{\infty}
{\ln\pars{x^{\mu} + 1} - \mu\ln\pars{x}\over \pars{x^{2} + 1}\ln\pars{x}}\,\dd x
\\[5mm] \implies &\
\bbx{\int_{0}^{\infty}
{\ln\pars{1 + x^{\mu}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x =
{1 \over 2}\mu\int_{0}^{\infty}{\dd x \over 1 + x^{2}} = {1 \over 4}\,\mu\pi}
\label{2}\tag{2}
\end{align}
\eqref{1} y \eqref{2} conducir a
$$
\bbx{\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}}
\,{\dd x \\pars{1 + x^{2}}\ln\pars{x}} =
{1 \over 4}\,11\pi - {1 \over 4}\,3\pi = {\2\pi}}
$$