The value of the integral ∫ _ - _ e 2 ^ _ e 2 e ^ x ( _ e ( e ^ x +...

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

Let us substitute ${e}^{x} = t$

$⇒ {e}^{x}dx = dt$

Lower limit $= {e}^{- \log_{e}\left(2\right)} = \dfrac{1}{2}$

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

$⇒ I = {∫}_{\dfrac{1}{2}}^{2}1 \times \log_{e}\left[t + \sqrt{1 + {t}^{2}}\right]dt$

Apply integration by-parts.

$∫ u\left(t\right)v\left(t\right)dt = u\left(t\right) ∫ v\left(t\right)dt - ∫ \left(u ' \left(t\right) ∫ v\left(t\right)dt\right)dt$

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

$⇒ u ' \left(t\right) = \dfrac{1}{t + \sqrt{1 + {t}^{2}}}\left(1 + \dfrac{2t}{2\sqrt{1 + {t}^{2}}}\right) = t$

$= {\left[tln\left(\sqrt{{t}^{2} + 1} + x\right)\right]}_{\dfrac{1}{2}}^{2} - {∫}_{1/2}^{2}\dfrac{t}{\sqrt{{t}^{2} + 1}}dt$

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

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

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

Hence this is the correct option.