Answer
$$\frac{1}{2}\ln \left( {4 + {e^{2x}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{2x}}}}{{4 + {e^{2x}}}}dx} \cr
& {\text{substitute }}u = 4 + {e^{2x}},{\text{ }}du = 2{e^{2x}}dx \cr
& \int {\frac{{{e^{2x}}}}{{4 + {e^{2x}}}}dx} = \int {\frac{{\left( {1/2} \right)}}{u}du} \cr
& = \frac{1}{2}\int {\frac{1}{u}} du \cr
& {\text{find the antiderivative}} \cr
& = \frac{1}{2}\ln \left| u \right| + C \cr
& {\text{with }}u = 4 + {e^{2x}} \cr
& = \frac{1}{2}\ln \left| {4 + {e^{2x}}} \right| + C \cr
& = \frac{1}{2}\ln \left( {4 + {e^{2x}}} \right) + C \cr} $$
