Answer
$$\ln \sqrt {\frac{{{e^4} + 1}}{2}} $$
Work Step by Step
$$\eqalign{
& \int_0^2 {\frac{{{e^{2x}}}}{{{e^{2x}} + 1}}} dx \cr
& {\text{Integrate using the substitution method}} \cr
& u = {e^{2x}} + 1,{\text{ }}du = 2{e^{2x}}dx,{\text{ }}dx = \frac{1}{{2{e^{2x}}}}du \cr
& {\text{The new limits of integration are}} \cr
& x = 2 \to u = {e^4} + 1 \cr
& x = 0 \to u = {e^0} + 1 = 2 \cr
& {\text{Substitute and integrate}} \cr
& \int_0^2 {\frac{{{e^{2x}}}}{{{e^{2x}} + 1}}} dx = \int_2^{{e^4} + 1} {\frac{{{e^{2x}}}}{u}} \left( {\frac{1}{{2{e^{2x}}}}} \right)du \cr
& = \frac{1}{2}\int_2^{{e^4} + 1} {\frac{1}{u}} du \cr
& = \frac{1}{2}\left[ {\ln \left| u \right|} \right]_2^{{e^4} + 1} \cr
& = \frac{1}{2}\left[ {\ln \left( {{e^4} + 1} \right) - \ln 2} \right] \cr
& = \frac{1}{2}\ln \left( {\frac{{{e^4} + 1}}{2}} \right) \cr
& = \ln \sqrt {\frac{{{e^4} + 1}}{2}} \cr} $$
