The value of the integral ∫ - log e 2 log e 2 e x log e e x + 1 + e 2 x d x is equal to

Let $I={\int }_{-{\log }_{e}2}^{{\log }_{e}2}{e}^x\left[{\log }_e\left(e^x+\sqrt{1+e^{2x}}\right)\right]dx$

Let us substitute $e^x=t$

$\Rightarrow e^{x}dx=dt$

Lower limit $=e^{-{\log }_e\left(2\right)}=\frac{1}{2}$

Upper limit $=e^{{\log }_e\left(2\right)}=2$

$\Rightarrow I={\int }_{\frac{1}{2}}^{2}1\times {\log }_e\left[t+\sqrt{1+t^2}\right]dt$

Apply integration by-parts.

$\int u\left(t\right)v\left(t\right)dt=u\left(t\right)\int v\left(t\right)dt-\int \left(u\text{'}\left(t\right)\int v\left(t\right)dt\right)dt$

Take $u\left(t\right)={\log }_e\left[t+\sqrt{1+t^2}\right], v\left(t\right)=1$

$\Rightarrow u\text{'}\left(t\right)=\frac{1}{t+\sqrt{1+t^2}}\left(1+\frac{2t}{2\sqrt{1+t^2}}\right)=t$

$={\left[t\ln \left(\sqrt{t^2+1}+x\right)\right]}_{\frac{1}{2}}^2-{\int }_{1/2}^2\frac{t}{\sqrt{t^2+1}}dt$

$={\left[t\ln \sqrt{t^2+1}+t-\sqrt{t^2+1}\right]}_{\frac{1}{2}}^2$

$=\left[2\ln \left(\sqrt{5}+2\right)-\sqrt{5}\right]-\left[\frac{1}{2}\ln \left(\sqrt{\frac{5}{2}}+\frac{1}{2}\right)-\sqrt{\frac{5}{2}}\right]$

$=\ln \left(\frac{\sqrt{2}(2+\sqrt{5}{)}^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}$

Hence this is the correct option.