## Calculus: Early Transcendentals (2nd Edition)

$$\sqrt 3 - 1 - \frac{\pi }{{12}}$$
\eqalign{ & \int_{\sqrt 2 }^2 {\frac{{\sqrt {{x^2} - 1} }}{x}} dx \cr & {\text{substitute }}x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta d\theta \cr & \int {\frac{{\sqrt {{x^2} - 1} }}{x}} dx = \int {\frac{{\sqrt {{{\sec }^2}\theta - 1} }}{{\sec \theta }}} \left( {\sec \theta \tan \theta } \right)d\theta \cr & {\text{simplify}} \cr & = \int {\sqrt {{{\tan }^2}\theta } } \tan \theta d\theta \cr & = \int {{{\tan }^2}\theta } d\theta \cr & = \int {\left( {{{\sec }^2}\theta - 1} \right)d\theta } \cr & {\text{find the antiderivative}} \cr & = \tan \theta - \theta + C \cr & {\text{write in terms of }}x \cr & = \sqrt {{x^2} - 1} - {\sec ^{ - 1}}x + C \cr & \int_{\sqrt 2 }^2 {\frac{{\sqrt {{x^2} - 1} }}{x}} dx = \left[ {\sqrt {{x^2} - 1} - {{\sec }^{ - 1}}x} \right]_{\sqrt 2 }^2 \cr & {\text{evaluate limits}} \cr & = \left[ {\sqrt {{{\left( 2 \right)}^2} - 1} - {{\sec }^{ - 1}}2} \right] - \left[ {\sqrt {{{\left( {\sqrt 2 } \right)}^2} - 1} - {{\sec }^{ - 1}}\sqrt 2 } \right] \cr & = \sqrt 3 - \frac{\pi }{3} - 1 + \frac{\pi }{4} \cr & = \sqrt 3 - 1 - \frac{\pi }{{12}} \cr}