Answer
$$\frac{{\sqrt 3 - \sqrt 2 }}{2}$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt 2 }^2 {\frac{{dx}}{{{x^2}\sqrt {{x^2} - 1} }}} \cr
& {\text{substitute }}x = \sec \theta ,{\text{ }}\,\,\,dx = \sec \theta \tan \theta d\theta \,\,\,\,\theta = {\sec ^{ - 1}}x \cr
& \,\,\,x = 2 \to \theta = {\sec ^{ - 1}}\left( 2 \right) = \pi /3 \cr
& \,\,\,x = \sqrt 2 \to \theta = {\sec ^{ - 1}}\left( {\sqrt 2 } \right) = \pi /4 \cr
& \cr
& {\text{write the integral in terms of }}\theta \cr
& \int_{\sqrt 2 }^2 {\frac{{dx}}{{{x^2}\sqrt {{x^2} - 1} }}} = \int_{\pi /4}^{\pi /3} {\frac{{\sec \theta \tan \theta }}{{{{\sec }^2}\theta \sqrt {{{\sec }^2}\theta - 1} }}d\theta } \cr
& \cr
& {\text{use the pythagorean identity }}{\tan ^2} + 1 = {\sec ^2}\theta \cr
& = \int_{\pi /4}^{\pi /3} {\frac{{\sec \theta \tan \theta }}{{{{\sec }^2}\theta \tan \theta }}d\theta } \cr
& = \int_{\pi /4}^{\pi /3} {\frac{1}{{\sec \theta }}d\theta } \cr
& = \int_{\pi /4}^{\pi /3} {\cos \theta d\theta } \cr
& = \left[ {\sin \theta } \right]_{\pi /4}^{\pi /3} \cr
& \cr
& {\text{evaluate the limits}} \cr
& = \sin \left( {\frac{\pi }{3}} \right) - \sin \left( {\frac{\pi }{4}} \right) \cr
& = \frac{{\sqrt 3 }}{2} - \frac{{\sqrt 2 }}{2} \cr
& = \frac{{\sqrt 3 - \sqrt 2 }}{2} \cr} $$
