If ${\int }_0^1 \frac{\sqrt{e^x}}{\sqrt{e^x+e^{-x}}}\mathrm{d}x=\log \frac{e+A}{1+\sqrt{2}}$ then $A$ is equal to 

${\int }_0^1 \frac{\sqrt{e^x}}{\sqrt{e^x+e^{-x}}}\mathrm{d}x={\int }_0^1 \frac{e^x}{\sqrt{e^{2x}+1}}\mathrm{d}x.$ Putting $e^x=t$

the last integral is equal to 

${\int }_1^e \frac{dt}{\sqrt{t^2+1}}=\log \mid t+\sqrt{t^2+1}{\parallel }_1^e$

        $\begin{array}{l}=\log \left(e+\sqrt{e^2+1}\right)-\log (1+\sqrt{2})\\ =\log \frac{e+\sqrt{e^2+1}}{1+\sqrt{2}}\end{array}$

So  $A=\sqrt{e^2+1}$