Answer
$$ - {\operatorname{sech} ^{ - 1}}{e^x} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {1 - {e^{2x}}} }}} \cr
& = \int {\frac{{dx}}{{\sqrt {1 - {{\left( {{e^x}} \right)}^2}} }}} \cr
& {\text{substitute }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& = \int {\frac{{dx}}{{\sqrt {1 - {{\left( {{e^x}} \right)}^2}} }}} = \int {\frac{{du/{e^x}}}{{\sqrt {1 - {u^2}} }}} \cr
& = \int {\frac{{du}}{{u\sqrt {1 - {u^2}} }}} \cr
& {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.6 }}\left( {{\text{see page 480}}} \right) \cr
& = - {\operatorname{sech} ^{ - 1}}\left| u \right| + C \cr
& {\text{write in terms of }}x \cr
& = - {\operatorname{sech} ^{ - 1}}\left| {{e^x}} \right| + C \cr
& = - {\operatorname{sech} ^{ - 1}}{e^x} + C \cr} $$
