Answer
$2(tan^{-1}{\sqrt e}-\frac{\pi}{4})$
Work Step by Step
the average value is
= $\frac{1}{\frac{1}{2}-0}\int_0^{\frac{1}{2}}(\frac{e^{x}}{1+e^{2x}})dx$
let
$u$ = $e^{x}$
$du$ = $e^{x}dx$
so
$\frac{1}{\frac{1}{2}-0}\int_0^{\frac{1}{2}}(\frac{e^{x}}{1+e^{2x}})dx$ = $2\int_1^{\sqrt e}\frac{du}{1+xu^{2}}$ =$2tan^{-1}u|_1^{\sqrt e}$ = $2(tan^{-1}{\sqrt e}-\frac{\pi}{4})$
