5. The value of the integral \( \int_{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x \) is equal to [Main April 11, 2023 (I)]
(a) \( \log _{e}\left(\frac{2(2+\sqrt{5})}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2} \)
(b) \( \log _{e}\left(\frac{\sqrt{2}(3-\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2} \)
(c) \( \log _{e}\left(\frac{(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2} \)
(d) \( \log _{e}\left(\frac{\sqrt{2}(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2} \)
