#### Answer

$${\sinh ^{ - 1}}\frac{{{e^x}}}{2} + C$$

#### Work Step by Step

$$\eqalign{
& \int {\frac{{{e^x}}}{{\sqrt {{e^{2x}} + 4} }}} dx \cr
& {\text{rewriting radicand}} \cr
& \int {\frac{{{e^x}}}{{\sqrt {{{\left( {{e^x}} \right)}^2} + {{\left( 2 \right)}^2}} }}} dx \cr
& u = {e^x},{\text{ }}du = {e^x}dx \cr
& {\text{find the antiderivative using the theorem 6}}{\text{.12}}{\text{, formula 2}} \cr
& \int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }}} = {\sinh ^{ - 1}}\frac{u}{a} + C \cr
& so \cr
& = {\sinh ^{ - 1}}\frac{{{e^x}}}{2} + C \cr} $$