Answer
$$2 - \frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt 2 }^2 {\frac{{\sqrt {2{x^2} - 4} }}{x}} dx \cr
& = \sqrt 2 \int_{\sqrt 2 }^2 {\frac{{\sqrt {{x^2} - 2} }}{x}} dx \cr
& {\text{substitute }}x = \sqrt 2 \sec \theta ,{\text{ }}\,\,\,dx = \sqrt 2 \sec \theta \tan \theta d\theta \,\,\,\,\theta = {\sec ^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) \cr
& \,\,\,x = 2 \to \theta = {\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 2 }}} \right) = \pi /4 \cr
& \,\,\,x = \sqrt 2 \to \theta = {\sec ^{ - 1}}\left( {\frac{{\sqrt 2 }}{{\sqrt 2 }}} \right) = 0 \cr
& \cr
& {\text{write the integral in terms of }}\theta \cr
& \sqrt 2 \int_{\sqrt 2 }^2 {\frac{{\sqrt {{x^2} - 2} }}{x}} dx = \sqrt 2 \int_0^{\pi /4} {\frac{{\sqrt {2{{\sec }^2}\theta - 2} }}{{\sqrt 2 \sec \theta }}} \left( {\sqrt 2 \sec \theta \tan \theta } \right)d\theta \cr
& = 2\int_0^{\pi /4} {\sqrt {{{\sec }^2}\theta - 1} \tan \theta } d\theta \cr
& \cr
& {\text{use the pythagorean identity }}{\tan ^2} + 1 = {\sec ^2}\theta \cr
& = 2\int_0^{\pi /4} {\sqrt {{{\tan }^2}\theta } \tan \theta } d\theta \cr
& = 2\int_0^{\pi /4} {{{\tan }^2}\theta } d\theta \cr
& = 2\int_0^{\pi /4} {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& = 2\left[ {\tan \theta - \theta } \right]_0^{\pi /4} \cr
& \cr
& {\text{evaluate the limits}} \cr
& = 2\left[ {\tan \left( {\frac{\pi }{4}} \right) - \frac{\pi }{4}} \right] - 2\left[ {\tan \left( 0 \right) - 0} \right] \cr
& = 2 - \frac{\pi }{2} \cr} $$
