Answer
$tan^{-1}~(e)-\frac{\pi}{4}$
Work Step by Step
$\int_{0}^{1}\frac{e^x}{1+e^{2x}}~dx$
Let $u = e^x$
$\frac{du}{dx} = e^x$
$dx = \frac{du}{e^x}$
When $x=0,$ then $u = e^0 = 1$
When $x=1,$ then $u = e^1 = e$
$\int_{1}^{e}\frac{e^x}{1+u^2}\frac{du}{e^x}$
$= \int_{1}^{e}\frac{du}{1+u^2}$
$= tan^{-1}~u\Big\vert_{1}^{e}$
$= tan^{-1}~(e) - tan^{-1}~(1)$
$= tan^{-1}~(e)-\frac{\pi}{4}$
