$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,}
\newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
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\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}}
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\begin{align}
&-\int_{0}^{\infty}\ln\pars{1 - \cosh\pars{x}}\,{x^{2} \over \expo{x}}\,\dd x =
-\int_{0}^{\infty}\bracks{\ln\pars{\cosh\pars{x} - 1} + \ic\pi}\,
x^{2}\expo{-x}\,\dd x
\\[5mm] = &\
-2\pi\ic - \int_{0}^{\infty}\ln\pars{{\expo{x} + \expo{-x} \over 2} - 1}
\,x^{2}\expo{-x}\,\dd x
\\[5mm] \stackrel{\substack{x\ =\ -\ln\pars{t}\\[0.25mm] t\ =\ \expo{-x}}\\ }{=}\,\,\,&
-2\pi\ic - \int_{1}^{0}\ln\pars{{1/t + t \over 2} - 1}\ln^{2}\pars{t}\,
t\,\,{\phantom{-}\dd t \over -t}
\\[5mm] = &\
-2\pi\ic - \int_{0}^{1}\ln\pars{\bracks{1 - t}^{2} \over 2t}\ln^{2}\pars{t}\,
\dd t
\\[5mm] = &\
-2\pi\ic - 2\
\underbrace{\int_{0}^{1}\ln^{2}\pars{t}\ln\pars{1 - t}\,\dd t}
_{\ds{-6 + {\pi^{2} \over 3} + 2\,\zeta\pars{3}}}\ +\
\ln\pars{2}\ \underbrace{\int_{0}^{1}\ln^{2}\pars{t}\,\dd t}_{\ds{2}}\ +\
\underbrace{\int_{0}^{1}\ln^{3}\pars{t}\,\dd t}_{\ds{-6}}\label{1}\tag{1}
\\[5mm] = &\
\bbx{6 + 2\ln\pars{2} - {2\pi^{2} \over 3} - 4\zeta\pars{3} - 2\pi\ic}
\approx -4.0015 + 6.2832\,\ic
\end{align}
Tenga en cuenta que las integrales en \eqref{1} se evalúan de la siguiente manera:
$$
\left\{\begin{array}{rcl}
\ds{\int_{0}^{1}\ln^{2}\pars{t}\ln\pars{1 - t}\,\dd t} & \ds{=} &
\ds{\left.\partiald[2]{}{\mu}\partiald{}{\nu}
{\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over
\Gamma\pars{\mu + \nu + 2}}\right\vert_{\ \mu =\ \nu\ =\ 0}}
\\[5mm]
\ds{\int_{0}^{1}\ln^{k}\pars{t}\,\dd t} & \ds{=} &
\ds{\left.\partiald[k]{}{\mu}\int_{0}^{1}t^{\mu}\,\dd t\,
\right\vert_{\ \mu\ =\ 0} = \pars{-1}^{k}\, k!}
\end{array}\right.
$$