Answer
$$\sqrt 3 - 1$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt 2 }^2 {\frac{{{{\sec }^2}\left( {{{\sec }^{ - 1}}x} \right)dx}}{{x\sqrt {{x^2} - 1} }}} \cr
& {\text{use the substitution method}}{\text{:}} \cr
& u = {\sec ^{ - 1}}x,{\text{ so that }}du = \frac{1}{{x\sqrt {{x^2} - 1} }}dx \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 2,{\text{ then }}u = {\sec ^{ - 1}}2 = \pi /3 \cr
& \,\,\,\,\,\,{\text{If }}x = \sqrt 2,{\text{ then }}u = {\sec ^{ - 1}}\left( {\sqrt 2 } \right) = \pi /4 \cr
& {\text{then}} \cr
& \int_{\sqrt 2 }^2 {\frac{{{{\sec }^2}\left( {{{\sec }^{ - 1}}x} \right)dx}}{{x\sqrt {{x^2} - 1} }}} = \int_{\pi /4}^{\pi /3} {{{\sec }^2}u} du \cr
& {\text{integrate}} \cr
& = \left( {\tan u} \right)_{\pi /4}^{\pi /3} \cr
& = \tan \left( {\frac{\pi }{3}} \right) - \tan \left( {\frac{\pi }{4}} \right) \cr
& {\text{simplify}} \cr
& = \sqrt 3 - 1 \cr} $$
