# Evaluate the following definite integrals. int_5^{3sqrt5}frac{dx}{sqrt{x^2-9}}

Evaluate the following definite integrals.
${\int }_{5}^{3\sqrt{5}}\frac{dx}{\sqrt{{x}^{2}-9}}$
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$\text{Consider x=3 sec(u)}$

${\int }_{5}^{3\sqrt{5}}\frac{dx}{\sqrt{{x}^{2}-9}}={\int }_{{\mathrm{sec}}^{-1}\left(\frac{5}{4}\right)}^{{\mathrm{sec}}^{-1}\left(\sqrt{5}\right)}\mathrm{sec}\left(u\right)du$
$=\left[\mathrm{ln}\left(\mathrm{tan}u+\mathrm{sec}u\right){\right]}_{{\mathrm{sec}}^{-1}\left(\frac{5}{4}\right)}^{{\mathrm{sec}}^{-1}\left(\sqrt{5}\right)}$
$=\mathrm{ln}\left(\mathrm{tan}\left({\mathrm{sec}}^{-1}\left(\sqrt{5}\right)\right)+\mathrm{sec}\left({\mathrm{sec}}^{-1}\left(\sqrt{5}\right)\right)\right)-\mathrm{ln}\left(\mathrm{tan}\left({\mathrm{sec}}^{-1}\left(\frac{5}{3}\right)\right)+\mathrm{sec}\left({\mathrm{sec}}^{-1}\left(\frac{5}{3}\right)\right)\right)$
$=\mathrm{ln}\left(2+\sqrt{5}\right)-\mathrm{ln}3$
$=0.34502$