Answer
$$\frac{\pi }{{12}}$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt 2 }^2 {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} \cr
& {\text{integrate using the rule for inverse trigonometric functions}} \cr
& = \left[ {{{\sec }^{ - 1}}x} \right]_{\sqrt 2 }^2 \cr
& {\text{using the fundamental theorem}} \cr
& = {\sec ^{ - 1}}\left( 2 \right) - {\sec ^{ - 1}}\left( {\sqrt 2 } \right) \cr
& {\text{simplify}} \cr
& = \frac{\pi }{3} - \frac{\pi }{4} \cr
& = \frac{\pi }{{12}} \cr} $$