Chapter 8.4, Problem 79E

Show that ${\displaystyle \int \frac{dx}{x\sqrt{x^2-1}}}=\{\begin{array}{ll}{\sec }^{-1}x+C={\tan }^{-1}\sqrt{x^2-1}+C\hfill & \text{if}x>1\hfill \\ -{\sec }^{-1}x+C=-{\tan }^{-1}\sqrt{x^2-1}+C\hfill & \text{if}x<-1.\hfill \end{array}$