Answer
$$\frac{1}{{10}}{\sec ^{ - 1}}\left( {\frac{y}{5}} \right) - \frac{{\sqrt {{y^2} - 25} }}{{2{y^2}}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {{y^2} - 25} }}{{{y^3}}}} dy \cr
& = \int {\frac{{\sqrt {{y^2} - {5^2}} }}{{{y^3}}}} dy \cr
& {\text{We set}}{\text{, }}y = 5\sec \theta,{\text{ }}dy = 5\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( {5\sec \theta } \right)}^2} - {5^2}} }}{{{{\left( {5\sec \theta } \right)}^3}}}} \left( {5\sec \theta \tan \theta } \right)d\theta \cr
& = \int {\frac{{\sqrt {25{{\sec }^2}\theta - 25} }}{{7\sec \theta }}} \left( {5\sec \theta \tan \theta } \right)d\theta \cr
& = \int {\frac{{\sqrt {25\left( {{{\sec }^2}\theta - 1} \right)} }}{{{5^3}{{\sec }^3}\theta }}} \left( {5\sec \theta \tan \theta } \right)d\theta \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{\sqrt {25{{\tan }^2}\theta } }}{{{5^3}{{\sec }^3}\theta }}} \left( {5\sec \theta \tan \theta } \right)d\theta \cr
& {\text{simplifying}} \cr
& = \int {\frac{{5\tan \theta }}{{{5^3}{{\sec }^3}\theta }}} \left( {5\sec \theta \tan \theta } \right)d\theta \cr
& = \frac{1}{5}\int {\frac{{{{\tan }^2}\theta }}{{{{\sec }^2}\theta }}} d\theta \cr
& = \frac{1}{5}\int {\left( {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)\left( {{{\cos }^2}\theta } \right)} d\theta \cr
& = \frac{1}{5}\int {{{\sin }^2}\theta } d\theta \cr
& = \frac{1}{5}\int {\left( {\frac{1}{2}\left( {1 - \cos 2\theta } \right)} \right)} d\theta \cr
& = \frac{1}{{10}}\int {\left( {1 - \cos 2\theta } \right)} d\theta \cr
& {\text{integrate}} \cr
& = \frac{1}{{10}}\theta - \frac{1}{{20}}\sin 2\theta + C \cr
& = \frac{1}{{10}}\theta - \frac{1}{{10}}\sin \theta \cos \theta + C \cr
& {\text{with sin }}\theta = \frac{{\sqrt {{y^2} - 25} }}{y}{\text{ and }}\cos \theta = \frac{5}{y} \cr
& = \frac{1}{{10}}{\sec ^{ - 1}}\left( {\frac{y}{5}} \right) - \frac{1}{{10}}\left( {\frac{{\sqrt {{y^2} - 25} }}{y}} \right)\left( {\frac{5}{y}} \right) + C \cr
& = \frac{1}{{10}}{\sec ^{ - 1}}\left( {\frac{y}{5}} \right) - \frac{{\sqrt {{y^2} - 25} }}{{2{y^2}}} + C \cr} $$
