The value of the integral ∫_(-logₑ 2)^(logₑ 2) e^x ( log ( e^x + √(1 + e^2x))) dx is equal to:

Question

The value of the integral $$\int_{-\log_e 2}^{\log_e 2} e^x \left( \log \left( e^x + \sqrt{1 + e^{2x}} \right) \right) dx$$ is equal to:

Accepted Answer

A. $$\log_e \left( \frac{\sqrt{2} \left( 2 + \sqrt{5} \right)^2}{\sqrt{1 + \sqrt{5}}} \right) - \frac{\sqrt{5}}{2}$$

Answer

B. $$\log_e \left( \frac{2 \left( 2 + \sqrt{5} \right)^2}{\sqrt{1 + \sqrt{5}}} \right) - \frac{\sqrt{5}}{2}$$

Answer

C. $$\log_e \left( \frac{\sqrt{2} \left( 3 - \sqrt{5} \right)^2}{\sqrt{1 + \sqrt{5}}} \right) + \frac{\sqrt{5}}{2}$$

Answer

D. $$\log_e \left( \frac{\left( 2 + \sqrt{5} \right)^2}{\sqrt{1 + \sqrt{5}}} \right) + \frac{\sqrt{5}}{2}$$

Answer & step-by-step solution