## Calculus: Early Transcendentals (2nd Edition)

$${\sec ^{ - 1}}x + \frac{{\sqrt {{x^2} - 1} }}{{2{x^2}}} + C$$
\eqalign{ & \int {\frac{{dx}}{{{x^3}\sqrt {{x^2} - 1} }}} \cr & {\text{substitute }}x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta d\theta \cr & \int {\frac{{dx}}{{{x^3}\sqrt {{x^2} - 1} }}} = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {\sec \theta } \right)}^3}\sqrt {{{\sec }^2}\theta - 1} }}} \cr & {\text{simplify}} \cr & = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {\sec \theta } \right)}^3}\sqrt {{{\sec }^2}\theta - 1} }}} \cr & {\text{pythagorean identity}} \cr & = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {\sec \theta } \right)}^3}\sqrt {{{\tan }^2}\theta } }}} \cr & = \int {\frac{1}{{{{\sec }^2}\theta }}d\theta } = \int {{{\cos }^2}\theta d\theta } \cr & {\text{half - angle formula}} \cr & = \int {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)d\theta } \cr & {\text{integrating}} \cr & = \theta + \frac{1}{4}\sin 2\theta + C \cr & = \theta + \frac{1}{2}\sin \theta \cos \theta + C \cr & = {\sec ^{ - 1}}x + \frac{1}{2}\left( {\frac{{\sqrt {{x^2} - 1} }}{x}} \right)\left( {\frac{1}{x}} \right) + C \cr & {\text{simplify}} \cr & = {\sec ^{ - 1}}x + \frac{{\sqrt {{x^2} - 1} }}{{2{x^2}}} + C \cr}