The value of integral \( \displaystyle \int_{-\log_e 2}^{\log_e 2} e^x \Big( \log_e( e^x + \sqrt{1 + e^{2x}} ) \Big) \, dx \) is equal to:
\( \log_e\!\left( \dfrac{\sqrt{2}(3-\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) + \dfrac{\sqrt{5}}{2} \)
\( \log_e\!\left( \dfrac{2(2+\sqrt{5})}{\sqrt{1+\sqrt{5}}} \right) - \dfrac{\sqrt{5}}{2} \)
\( \log_e \left( \dfrac{\sqrt{2}(2 + \sqrt{5})^2}{\sqrt{1 + \sqrt{5}}} \right) - \dfrac{\sqrt{5}}{2} \)
\( \log_e\!\left( \dfrac{(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) + \dfrac{\sqrt{5}}{2} \)