Answer
$\dfrac{-1}{8} \ln (\dfrac{2}{5}) \approx 0.1145$
Work Step by Step
Given: $I=\int_0^{ \ln 2} \dfrac{e^{-2x}}{1+4 e^{-2x}} \ dx$
Let us consider that $u=1+4 e^{-2x} \implies dx=\dfrac{-du}{8e^{-2x}}$
Now, we have $I=\int_0^{ \ln 2} \dfrac{e^{-2x}}{u} (\dfrac{-du}{8e^{-2x}})$
or, $=-\dfrac{1}{8} \int_0^{ \ln 2} \dfrac{\ du}{u}$
or, $=\dfrac{-1}{8}[\ln |u|]_0^{ \ln 2}$
or, $=\dfrac{-1}{8}[\ln |1+4 e^{-2x} |]_0^{ \ln 2}$
or, $=\dfrac{-1}{8}[\ln |1+4 e^{-2(\ln 2)} |]+\dfrac{-1}{8}[\ln |1+4 e^{-2(0)} |]$
or, $=\dfrac{-1}{8} \ln (\dfrac{2}{5}) \approx 0.1145$
