The value of the integral $\underset{-{\log }_{\mathrm{e}}2}{\overset{{\log }_{\mathrm{e}}2}{\int }}{\mathrm{e}}^{\mathrm{x}}\left({\log }_{\mathrm{e}}({\mathrm{e}}^{\mathrm{x}}+\sqrt{1+{\mathrm{e}}^{2\mathrm{x}}})\right)\mathrm{\hspace{0.17em}}\mathrm{dx}$

$\mathrm{I}=\underset{-{{\log }_{\mathrm{e}}}^2}{\overset{{{\log }_{\mathrm{e}}}^2}{\int }}{\mathrm{e}}^{\mathrm{x}}{\log }_{\mathrm{e}}({\mathrm{e}}^{\mathrm{x}}+\sqrt{1+{\mathrm{e}}^{2\mathrm{x}}})\mathrm{dx}$
$\mathrm{let}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}\Rightarrow {\mathrm{e}}^{\mathrm{x}}.\mathrm{dx}=\mathrm{dt}$              $\begin{array}{l}\mathrm{x}:-{{\log }_{\mathrm{e}}}^2\Rightarrow \mathrm{t}:\frac{1}{2}\\ \mathrm{x}:{{\log }_{\mathrm{e}}}^2\to \mathrm{t}:\to 2\end{array}$
$\mathrm{I}=\underset{1/2}{\overset{2}{\int }}{\log }_{\mathrm{e}}(\mathrm{t}+\sqrt{1+{\mathrm{t}}^2})\mathrm{dt}$           $\Rightarrow I={\mathrm{t}.{\log }_{\mathrm{e}}(\mathrm{t}+\sqrt{1+{\mathrm{t}}^2})|}_{\frac{1}{2}}^2-\underset{\frac{1}{2}}{\overset{2}{\int }}\frac{\mathrm{t}}{\sqrt{1+{\mathrm{t}}^2}}\mathrm{dt}$
$$\begin{gathered} I={2{{\log }_{\mathrm{e}}}^{(2+\sqrt{5})}-\frac{1}{2}\log \left(\frac{1+\sqrt{5}}{2}\right)-\left(\sqrt{1+{\mathrm{t}}^2}\right)|}_{\frac{1}{2}}^2 \\ {I={\log }_{\mathrm{e}}}^{{(2+\sqrt{5})}^2}+\log \frac{\sqrt{2}}{\sqrt{1+\sqrt{5}}}-(\sqrt{5}-\frac{\sqrt{5}}{2}) \\ I=\log \frac{\sqrt{2}.{(2+\sqrt{5})}^2}{\sqrt{1+\sqrt{5}}}-\frac{\sqrt{5}}{2} \end{gathered}$$