Answer
$${\sin ^{ - 1}}\left( {{e^x}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{\sqrt {1 - {e^{2x}}} }}} dx \cr
& {\text{substitute }}u = {e^x},{\text{ }} \cr
& du = {e^x}dx \cr
& = \int {\frac{{{e^x}}}{{\sqrt {1 - {{\left( {{e^x}} \right)}^2}} }}} dx \cr
& = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& {\text{find the antiderivative }}\left( {{\text{formula 21 page 489}}} \right) \cr
& = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} = {\sin ^{ - 1}}\left( {\frac{u}{1}} \right) + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = {e^x} \cr
& = {\sin ^{ - 1}}\left( {{e^x}} \right) + C \cr} $$
