$\int \dfrac{e^{-x}}{1+e^{2x}}dx =\int \dfrac{1}{e^x(1+e^{2x})}dx=\int \dfrac{e^x dx}{e^{2x}(1+e^{2x})} =\int \dfrac{1}{t^2(1+t^2)}dt$
$=\int \bigg (\dfrac{1}{t^2}-\dfrac{1}{t^2+1} \bigg ) dt =-\dfrac{1}{t}-arc \tan t=-\dfrac{1}{e^{-x}}-arc \tan (e^{-x})+C$