Answer
$$F'\left( x \right) = 2{e^{2x}}\ln \left( {{e^{2x}} + 1} \right)$$
Work Step by Step
$$\eqalign{
& F\left( x \right) = \int_0^{{e^{2x}}} {\ln \left( {t + 1} \right)} dt \cr
& {\text{Differentiate both sides with respect to }}x \cr
& F'\left( x \right) = \frac{d}{{dx}}\left[ {\int_0^{{e^{2x}}} {\ln \left( {t + 1} \right)} dt} \right] \cr
& {\text{Use the second fundamental theorem of calculus }} \cr
& {\text{and the chain rule }}\left( {{\text{see page 284}}} \right) \cr
& F'\left( x \right) = \ln \left( {{e^{2x}} + 1} \right)\frac{d}{{dx}}\left[ {{e^{2x}}} \right] \cr
& F'\left( x \right) = \ln \left( {{e^{2x}} + 1} \right)\left( {2{e^{2x}}} \right) \cr
& {\text{Simplifying}} \cr
& F'\left( x \right) = 2{e^{2x}}\ln \left( {{e^{2x}} + 1} \right) \cr} $$
