Answer
$$\ln 3$$
Work Step by Step
$$\eqalign{
& \int_3^5 {\frac{1}{{\sqrt {{x^2} - 9} }}} dx \cr
& \frac{1}{{\sqrt {36 - {x^2}} }}{\text{ has an infinite discontinuity at }}x = 3,{\text{ so we can write}} \cr
& \int_3^5 {\frac{1}{{\sqrt {{x^2} - 9} }}} dx = \mathop {\lim }\limits_{b \to {3^ + }} \int_b^5 {\frac{1}{{\sqrt {{x^2} - 9} }}} dx \cr
& {\text{Integrate by tables, use }}\int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}} dx = \ln \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C \cr
& \mathop {\lim }\limits_{b \to {3^ + }} \int_b^5 {\frac{1}{{\sqrt {{x^2} - 9} }}} d\theta = \mathop {\lim }\limits_{b \to {3^ + }} \left[ {\ln \left| {x + \sqrt {{x^2} - 9} } \right|} \right]_b^5 \cr
& = \mathop {\lim }\limits_{b \to {3^ + }} \left[ {\ln \left| {5 + \sqrt {{5^2} - 9} } \right| - \ln \left| {b + \sqrt {{b^2} - 9} } \right|} \right] \cr
& = \mathop {\lim }\limits_{b \to {3^ + }} \left[ {\ln \left| {5 + 4} \right| - \ln \left| {b + \sqrt {{b^2} - 9} } \right|} \right] \cr
& = \mathop {\lim }\limits_{b \to {3^ + }} \left[ {\ln 9 - \ln \left| {b + \sqrt {{b^2} - 9} } \right|} \right] \cr
& {\text{Evaluate the limit when }}b \to 6 \cr
& = \ln 9 - \ln \left| {3 + \sqrt {{3^2} - 9} } \right| \cr
& = \ln 9 - \ln 3 \cr
& = \ln 3 \cr
& {\text{The following graph confirms the result}} \cr} $$
