$I=\int \limits_{-\ln 2}^{\ln 2} e^x\left(\ln \left(e^x+\sqrt{1+e^{2 x}}\right)\right) d x$
Put $e ^x=t \Rightarrow e^x d x=d t$
$I=\int \limits_{1 / 2}^2 \ln \left(t+\sqrt{1+t^2}\right) d t$
Applying integration by parts.
$=\left[t \ln \left( t +\sqrt{1+ t ^2}\right)\right]_{\frac{1}{2}}^2-\int \limits_{1 / 2}^2 \frac{ t }{ t +\sqrt{1+ t ^2}}\left(1+\frac{2 t }{2 \sqrt{1+ t ^2}}\right) d t$
$=2 \ln (2+\sqrt{5})-\frac{1}{2} \ln \left(\frac{1+\sqrt{5}}{2}\right)-\int \limits_{1 / 2}^2 \frac{ t }{\sqrt{1+ t ^2}} dt$
$=2 \ln (2+\sqrt{5})-\frac{1}{2} \ln \left(\frac{1+\sqrt{5}}{2}\right)-\frac{\sqrt{5}}{2}$
$=\ln \left(\frac{(2+\sqrt{5})^2}{\left(\frac{\sqrt{5}+1}{2}\right)^{\frac{1}{2}}}\right)-\frac{\sqrt{5}}{2}$