Answer
$$\frac{1}{6}\left[ {\sqrt {48} - {{\sec }^{ - 1}}\left( 7 \right)} \right]$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \frac{{\sqrt {{x^2} - 1} }}{x},{\text{ on }}\underbrace {\left[ {1,7} \right]}_{\left[ {a,b} \right]} \cr
& {\text{The average value of }}f{\text{ on the interval }}\left[ {a,b} \right]{\text{ is}} \cr
& {f_{{\text{ave}}}} = \frac{1}{{b - a}}\int_a^b {f\left( t \right)} dt \cr
& {\text{Therefore,}} \cr
& {f_{{\text{ave}}}} = \frac{1}{{7 - 1}}\int_1^7 {\frac{{\sqrt {{x^2} - 1} }}{x}} dx \cr
& {f_{{\text{ave}}}} = \frac{1}{6}\int_1^7 {\frac{{\sqrt {{x^2} - 1} }}{x}} dx \cr
& {\text{Let }}x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta \cr
& {\text{The new limits of integration are:}} \cr
& x = 7 \to \theta = {\sec ^{ - 1}}\left( 7 \right) \cr
& x = 1 \to \theta = {\sec ^{ - 1}}\left( 1 \right) = 0 \cr
& {\text{Substituting}} \cr
& = \frac{1}{6}\int_{}^{} {\frac{{\sqrt {{{\sec }^2}\theta - 1} }}{{\sec \theta }}} \left( {\sec \theta \tan \theta } \right)d\theta \cr
& = \frac{1}{6}\int_{}^{} {\sqrt {{{\tan }^2}\theta } } \tan \theta d\theta \cr
& = \frac{1}{6}\int_{}^{} {{{\tan }^2}\theta } d\theta \cr
& = \frac{1}{6}\int_{}^{} {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{Integrating}} \cr
& = \frac{1}{6}\left( {\tan \theta - \theta } \right) + C \cr
& = \frac{1}{6}\left( {\sqrt {{x^2} - 1} - {{\sec }^{ - 1}}x} \right) + C \cr
& {\text{Therefore}} \cr
& {f_{{\text{ave}}}} = \frac{1}{6}\left[ {\sqrt {{x^2} - 1} - {{\sec }^{ - 1}}x} \right]_1^7 \cr
& {f_{{\text{ave}}}} = \frac{1}{6}\left[ {\sqrt {{7^2} - 1} - {{\sec }^{ - 1}}\left( 7 \right)} \right] - \frac{1}{6}\left[ {\sqrt {{1^2} - 1} - {{\sec }^{ - 1}}\left( 1 \right)} \right] \cr
& {f_{{\text{ave}}}} = \frac{1}{6}\left[ {\sqrt {48} - {{\sec }^{ - 1}}\left( 7 \right)} \right] - \frac{1}{6}\left[ {\sqrt 0 - 0} \right] \cr
& {f_{{\text{ave}}}} = \frac{1}{6}\left[ {\sqrt {48} - {{\sec }^{ - 1}}\left( 7 \right)} \right] \cr} $$
