Answer
$${\cosh ^{ - 1}}\frac{x}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {{x^2} - 9} }}} \cr
& {\text{rewriting }} \cr
& = \int {\frac{{dx}}{{\sqrt {{x^2} - {{\left( 3 \right)}^2}} }}} \cr
& {\text{find the antiderivative using the theorem 6}}{\text{.12}}{\text{, formula 1}} \cr
& \int {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} = {\cosh ^{ - 1}}\frac{x}{a} + C \cr
& {\text{ with }}a = 3,so \cr
& = {\cosh ^{ - 1}}\frac{x}{3} + C \cr} $$
