MathematicsMedium210×since 2002Q3464The value of the integral \int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x is equal to :Aloge((2+5)21+5)+52\log _{e}\left(\frac{(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}loge(1+5(2+5)2)+25Bloge(2(2+5)21+5)−52\log _{e}\left(\frac{\sqrt{2}(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}loge(1+52(2+5)2)−25Cloge(2(2+5)1+5)−52\log _{e}\left(\frac{2(2+\sqrt{5})}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}loge(1+52(2+5))−25Dloge(2(3−5)21+5)+52\log _{e}\left(\frac{\sqrt{2}(3-\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}loge(1+52(3−5)2)+25Check answerSkip