Answer
$$\ln \left( {{e^x} + \sqrt {{e^{2x}} + 4} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{\sqrt {4 + {e^{2x}}} }}dx} \cr
& {\text{substitute }}u = {e^x},\,\,du = {e^x}dx \cr
& \int {\frac{{{e^x}}}{{\sqrt {4 + {e^{2x}}} }}dx} = \int {\frac{{{e^x}}}{{\sqrt {{2^2} + {{\left( {{e^x}} \right)}^2}} }}dx} \cr
& = \int {\frac{{du}}{{\sqrt {{2^2} + {u^2}} }}} \cr
& {\text{find the antiderivative using the theorem 6}}{\text{.8}}{\text{.6 }}\left( {{\text{see page 480}}} \right) \cr
& \int {\frac{{du}}{{\sqrt {{a^2} + {u^2}} }} = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C} \cr
& = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = {e^x},{\text{ and }}a = 2 \cr
& = \ln \left( {{e^x} + \sqrt {{{\left( {{e^x}} \right)}^2} + {{\left( 2 \right)}^2}} } \right) + C \cr
& {\text{Simplify}} \cr
& = \ln \left( {{e^x} + \sqrt {{e^{2x}} + 4} } \right) + C \cr} $$
