If  $\int e^x\ln (e^x+\sqrt{1+e^{2x}})dx=\varphi \left(x\right)+c$ and if  $\varphi \left(0\right)=\ln \left(\frac{1}{(\sqrt{2}-1)e^{\sqrt{2}}}\right)$ then arithmetic mean of  $\varphi \left(\ln 2\right)$ and $\sqrt{5}$  is

$e^x=t$

$\Rightarrow \int \ln (t+\sqrt{t^2+1})dx=t\ln (t+\sqrt{t^2+1})-\int \frac{1}{\sqrt{t^2+1}}tdt$

$t\ln (t+\sqrt{t^2+1})-\sqrt{t^2+1}+c$

$=e^{x}ln(e^x+\sqrt{e^{2x}+1})-\sqrt{e^{2x}+1}+c$

$c=0$

Now$\varphi \left(\ln 2\right)=2\ln (2+\sqrt{5})-\sqrt{5}$

$\therefore \frac{\varphi \left(\ln 2\right)+\sqrt{5}}{2}=\ln (2+\sqrt{5})$