Answer
$$\frac{1}{2}\left( {\arcsin {e^x} + {e^x}\sqrt {1 - {e^{2x}}} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{e^x}\sqrt {1 - {e^{2x}}} dx} \cr
& = \int {{e^x}\sqrt {1 - {{\left( {{e^x}} \right)}^2}} dx} \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& \int {{e^x}\sqrt {1 - {{\left( {{e^x}} \right)}^2}} dx} = \int {\sqrt {1 - {u^2}} du} \cr
& {\text{Use the special integration formula }}\left( {{\text{Theorem 8}}{\text{.2}}} \right) \cr
& \int {\sqrt {{a^2} - {u^2}} du = \frac{1}{2}\left( {{a^2}\arcsin \frac{u}{a} + u\sqrt {{a^2} - {u^2}} } \right) + C} \cr
& \int {\sqrt {1 - {u^2}} du} = \frac{1}{2}\left( {{a^2}\arcsin \frac{u}{a} + u\sqrt {{a^2} - {u^2}} } \right) + C \cr
& = \frac{1}{2}\left( {\arcsin u + u\sqrt {1 - {u^2}} } \right) + C \cr
& {\text{Substitute }}u = {e^x} \cr
& = \frac{1}{2}\left( {\arcsin {e^x} + {e^x}\sqrt {1 - {{\left( {{e^x}} \right)}^2}} } \right) + C \cr
& = \frac{1}{2}\left( {\arcsin {e^x} + {e^x}\sqrt {1 - {e^{2x}}} } \right) + C \cr} $$
