Answer
$$6{\sec ^{ - 1}}\left| {\frac{y}{4}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{24dy}}{{y\sqrt {{y^2} - 16} }}} \cr
& = \int {\frac{{24dy}}{{y\sqrt {{y^2} - {4^2}} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = y,{\text{ so that }}du = dy \cr
& {\text{then}} \cr
& = \int {\frac{{24du}}{{u\sqrt {{u^2} - {4^2}} }}} \cr
& = 24\int {\frac{{du}}{{u\sqrt {{u^2} - {4^2}} }}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}se{c^{ - 1}}\left| {\frac{u}{a}} \right| + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 4 \cr
& = 24\left( {\frac{1}{4}{{\sec }^{ - 1}}\left| {\frac{u}{4}} \right|} \right) + C \cr
& = 6{\sec ^{ - 1}}\left| {\frac{u}{4}} \right| + C \cr
& {\text{replace }}y{\text{ for }}u \cr
& = 6{\sec ^{ - 1}}\left| {\frac{y}{4}} \right| + C \cr} $$
