\(\text{Given integrals is }\int_5^{3\sqrt5}\frac{dx}{\sqrt{x^2-9}}\)

\(\text{Consider x=3 sec(u)}\)

\(\text{Suvstitute x=3 sec(u) in }\int_5^{3\sqrt5}\frac{dx}{\sqrt{x^2-9}}\)

\(\int_5^{3\sqrt5}\frac{dx}{\sqrt{x^2-9}}=\int_{\sec^{-1} (\frac{5}{4})}^{\sec^{-1}(\sqrt5)} \sec(u)du\)

\(=[\ln(\tan u+\sec u)]_{\sec^{-1} (\frac{5}{4})}^{\sec^{-1}(\sqrt5)}\)

\(=\ln(\tan(\sec^{-1}(\sqrt5))+\sec(\sec^{-1}(\sqrt5)))-\ln(\tan(\sec^{-1}(\frac{5}{3}))+\sec(\sec^{-1}(\frac{5}{3})))\)

\(=\ln(2+\sqrt5)-\ln3\)

\(=0.34502\)

\(\text{Consider x=3 sec(u)}\)

\(\text{Suvstitute x=3 sec(u) in }\int_5^{3\sqrt5}\frac{dx}{\sqrt{x^2-9}}\)

\(\int_5^{3\sqrt5}\frac{dx}{\sqrt{x^2-9}}=\int_{\sec^{-1} (\frac{5}{4})}^{\sec^{-1}(\sqrt5)} \sec(u)du\)

\(=[\ln(\tan u+\sec u)]_{\sec^{-1} (\frac{5}{4})}^{\sec^{-1}(\sqrt5)}\)

\(=\ln(\tan(\sec^{-1}(\sqrt5))+\sec(\sec^{-1}(\sqrt5)))-\ln(\tan(\sec^{-1}(\frac{5}{3}))+\sec(\sec^{-1}(\frac{5}{3})))\)

\(=\ln(2+\sqrt5)-\ln3\)

\(=0.34502\)