Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises - Page 503: 54

$${\cosh ^{ - 1}}\frac{x}{4} + C$$

$$\eqalign{ & \int {\frac{{dx}}{{\sqrt {{x^2} - 16} }}} \cr & {\text{rewriting the denominator}} \cr & \int {\frac{{dx}}{{\sqrt {{x^2} - {{\left( 4 \right)}^2}} }}} \cr & {\text{find the antiderivative using the theorem 6}}{\text{.12}}{\text{, formula 1}} \cr & \int {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} = {\cosh ^{ - 1}}\frac{x}{a} + C \cr & ,so \cr & = {\cosh ^{ - 1}}\frac{x}{4} + C \cr} $$