The value of the integral $$\int_{-\log_e 2}^{\log_e 2} e^x \log_e e^x + \sqrt{1 + e^{2x}} \, dx$$ is equal to

We need to evaluate $$\int_{-\ln 2}^{\ln 2} e^x \cdot \ln\left(e^x + \sqrt{1 + e^{2x}}\right) dx$$.

Let $$t = e^x$$, so $$dt = e^x \, dx$$. When $$x = -\ln 2$$, $$t = \frac{1}{2}$$; when $$x = \ln 2$$, $$t = 2$$.

The integral becomes:

$$I = \int_{1/2}^{2} \ln\left(t + \sqrt{1 + t^2}\right) dt$$

Note that $$\ln(t + \sqrt{1+t^2}) = \sinh^{-1}(t)$$.

We integrate by parts: let $$u = \sinh^{-1}(t)$$ and $$dv = dt$$.

Then $$du = \frac{1}{\sqrt{1+t^2}} dt$$ and $$v = t$$.

$$I = \left[t \cdot \sinh^{-1}(t)\right]_{1/2}^{2} - \int_{1/2}^{2} \frac{t}{\sqrt{1+t^2}} dt$$

The second integral: $$\int \frac{t}{\sqrt{1+t^2}} dt = \sqrt{1+t^2} + C$$.

So: $$I = \left[t \cdot \sinh^{-1}(t) - \sqrt{1+t^2}\right]_{1/2}^{2}$$

At $$t = 2$$: $$2\ln(2 + \sqrt{5}) - \sqrt{5}$$

At $$t = \frac{1}{2}$$: $$\frac{1}{2}\ln\left(\frac{1}{2} + \sqrt{\frac{5}{4}}\right) - \frac{\sqrt{5}}{2} = \frac{1}{2}\ln\left(\frac{1+\sqrt{5}}{2}\right) - \frac{\sqrt{5}}{2}$$

Therefore:

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

$$= 2\ln(2+\sqrt{5}) - \frac{1}{2}\ln(1+\sqrt{5}) + \frac{1}{2}\ln 2 - \frac{\sqrt{5}}{2}$$

$$= \ln(2+\sqrt{5})^2 - \frac{1}{2}\ln(1+\sqrt{5}) + \frac{1}{2}\ln 2 - \frac{\sqrt{5}}{2}$$

$$= \ln(2+\sqrt{5})^2 - \ln\sqrt{1+\sqrt{5}} + \ln\sqrt{2} - \frac{\sqrt{5}}{2}$$

$$= \ln\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} - \frac{\sqrt{5}}{2}$$

The answer is Option A.