Logarithm and square root definite integral

First note that $\tfrac{(e-1)^2}{4e} = \sinh^2\tfrac12$ . With the substitution $x = \sinh^2u$ , we have $\begin{aligned} A & = \; \int_0^{\frac12} \ln(\cosh u + \sinh u) \times 2\sinh u \cosh u\,du \; = \; \int_0^{\frac12} u\sinh 2u\,du \; = \; \tfrac14\int_0^1 v \sinh v\,dv \\ & = \; \frac14\left(\Big[v\cosh v\Big]_0^1 - \int_0^1 \cosh v\,dv \right) \;= \; \frac14\Big[v \cosh v - \sinh v \Big]_0^1 \; = \; \tfrac14(\cosh 1 - \sinh 1) \\ & = \; \tfrac{1}{4e} \end{aligned}$ so that $eA = \boxed{\tfrac14}$ .