Answer
$$\frac{\pi }{{12}}$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt 2 }^2 {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} \cr
& {\text{integrate using the formula }}\int {\frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }} = {{\sec }^{ - 1}}u + C,{\text{ }}} \left( {{\text{see page 468}}} \right) \cr
& {\text{let }}a = 1,{\text{ u}} = x \cr
& \int_{\sqrt 2 }^2 {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} = \left. {\left( {{{\sec }^{ - 1}}x} \right)} \right|_{\sqrt 2 }^2 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = {\sec ^{ - 1}}\left( 2 \right) - {\sec ^{ - 1}}\left( {\sqrt 2 } \right) \cr
& or \cr
& = co{s^{ - 1}}\left( {\frac{1}{2}} \right) - {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) \cr
& {\text{simplify}} \cr
& = \frac{\pi }{3} - \frac{\pi }{4} \cr
& = \frac{{4\pi - 3\pi }}{{12}} \cr
& = \frac{\pi }{{12}} \cr} $$
