Answer
$$\sqrt {{y^2} - 49} - 7{\sec ^{ - 1}}\left( {\frac{y}{7}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {{y^2} - 49} }}{y}} dy \cr
& = \int {\frac{{\sqrt {{y^2} - {7^2}} }}{y}} dy \cr
& {\text{We set}}{\text{, }}y = 7\sec \theta,{\text{ }}dy = 7\sec \theta \tan \theta d\theta,{\text{ }}0 \leqslant \theta \leqslant \frac{\pi }{2} \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& = \int {\frac{{\sqrt {{{\left( {7\sec \theta } \right)}^2} - {7^2}} }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr
& = \int {\frac{{\sqrt {49{{\sec }^2}\theta - 49} }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr
& = \int {\frac{{\sqrt {49\left( {{{\sec }^2}\theta - 1} \right)} }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{\sqrt {49{{\tan }^2}\theta } }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr
& {\text{simplifying}} \cr
& = \int {\frac{{\sqrt {49{{\tan }^2}\theta } }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr
& = 7\int {{{\tan }^2}\theta } d\theta \cr
& = 7\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{integrate}} \cr
& = 7\tan \theta - 7\theta + C \cr
& {\text{with tan }}\theta = \frac{{\sqrt {{y^2} - 49} }}{7}{\text{ and }}\theta = {\sec ^{ - 1}}\left( {\frac{y}{7}} \right) \cr
& = 7\left( {\frac{{\sqrt {{y^2} - 49} }}{7}} \right) - 7{\sec ^{ - 1}}\left( {\frac{y}{7}} \right) + C \cr
& = \sqrt {{y^2} - 49} - 7{\sec ^{ - 1}}\left( {\frac{y}{7}} \right) + C \cr} $$
