Answer
$${\sec ^{ - 1}}\left| {2y} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dy}}{{y\sqrt {4{y^2} - 1} }}} \cr
& = \int {\frac{{dy}}{{y\sqrt {{{\left( {2y} \right)}^2} - 1} }}} \cr
& {\text{use the substitution method}}{\text{:}} \cr
& u = 2y,{\text{ so that }}du = 2dy \cr
& {\text{then}} \cr
& \int {\frac{{dy}}{{y\sqrt {{{\left( {2y} \right)}^2} - 1} }}} = \int {\frac{{du/2}}{{2y\sqrt {{u^2} - 1} }}} \cr
& = \int {\frac{{du}}{{u\sqrt {{u^2} - 1} }}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}se{c^{ - 1}}\left| {\frac{u}{a}} \right| + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = {\sec ^{ - 1}}\left| u \right| + C \cr
& {\text{replace }}2y{\text{ for }}u \cr
& = {\sec ^{ - 1}}\left| {2y} \right| + C \cr} $$
