Answer
$${\sec ^{ - 1}}\left( {{e^v}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dv}}{{\sqrt {{e^{2v}} - 1} }}} \cr
& {\text{write the integrand as}} \cr
& \int {\frac{{{e^v}dv}}{{{e^v}\sqrt {{e^{2v}} - 1} }}} \cr
& \cr
& {\text{Let }}u = {e^v},\,\,\,\,du = {e^v}dv \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\frac{{{e^v}dv}}{{{e^v}\sqrt {{e^{2v}} - 1} }}} = \int {\frac{{du}}{{u\sqrt {{u^2} - 1} }}} \cr
& \cr
& {\text{use the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}{{\sec }^{ - 1}}\left( {\frac{u}{a}} \right) + C} \cr
& = {\sec ^{ - 1}}\left( u \right) + C \cr
& \cr
& {\text{Write in terms of }}v;{\text{ substitute }}{e^v}{\text{ for }}u \cr
& = {\sec ^{ - 1}}\left( {{e^v}} \right) + C \cr} $$
