Answer
$$A = \frac{\pi }{{12}}$$
Work Step by Step
$$\eqalign{
& {\text{The area of the region is given by}} \cr
& A = \int_{2/\sqrt 2 }^2 {\frac{1}{{x\sqrt {{x^2} - 1} }}} dx \cr
& {\text{Integrate using basic integration rules}} \cr
& A = \left[ {{{\sec }^{ - 1}}x} \right]_{2/\sqrt 2 }^2 \cr
& A = {\sec ^{ - 1}}\left( 2 \right) - {\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 2 }}} \right) \cr
& A = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) - {\cos ^{ - 1}}\left( {\frac{{\sqrt 2 }}{2}} \right) \cr
& {\text{Simplify}} \cr
& A = \frac{\pi }{3} - \frac{\pi }{4} \cr
& A = \frac{\pi }{{12}} \cr} $$
