Answer
$$\ln \left( {\frac{{\sqrt 5 + 2}}{3}} \right)$$
Work Step by Step
$$\eqalign{
& \int_5^{3\sqrt 5 } {\frac{{dx}}{{\sqrt {{x^2} - 9} }}} \cr
& {\text{using the Integral formulas in the Theorem 6}}{\text{.12 }}\,\int {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }} = {{\cosh }^{ - 1}}\frac{x}{a} + C{\text{.}}} \cr
& {\text{then}}{\text{, set }}a = 3 \cr
& \int_5^{3\sqrt 5 } {\frac{{dx}}{{\sqrt {{x^2} - 9} }}} = \left( {{{\cosh }^{ - 1}}\frac{x}{3}} \right)_5^{3\sqrt 5 } \cr
& {\text{evaluate the limits of integration}} \cr
& = {\cosh ^{ - 1}}\frac{{3\sqrt 5 }}{3} - {\cosh ^{ - 1}}\frac{5}{3} \cr
& = {\cosh ^{ - 1}}\sqrt 5 - {\cosh ^{ - 1}}\frac{5}{3} \cr
& {\text{Using the theorem 6}}{\text{.10 to express the answer in terms of logarithms}}:{\text{ }} \cr
& {\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right){\text{ }}\left( {x \geqslant 1} \right){\text{ then}}{\text{,}} \cr
& {\cosh ^{ - 1}}\sqrt 5 - {\cosh ^{ - 1}}\frac{5}{3} = \ln \left( {\sqrt 5 + \sqrt {{{\left( {\sqrt 5 } \right)}^2} - 1} } \right) - \ln \left( {\frac{5}{3} + \sqrt {{{\left( {\frac{5}{3}} \right)}^2} - 1} } \right) \cr
& {\text{Simplifying}} \cr
& = \ln \left( {\sqrt 5 + \sqrt 4 } \right) - \ln \left( {\frac{5}{3} + \sqrt {\frac{{16}}{9}} } \right) \cr
& = \ln \left( {\sqrt 5 + 2} \right) - \ln \left( 3 \right) \cr
& = \ln \left( {\frac{{\sqrt 5 + 2}}{3}} \right) \cr} $$
