## Calculus: Early Transcendentals (2nd Edition)

$= \sqrt {{x^2} - 9} - 3\operatorname{arcsec} \,\left( {\frac{x}{3}} \right) + C$
$\begin{gathered} \int_{}^{} {\frac{{\sqrt {{x^2} - 9} }}{x}} dx\,\,\,,\,\,\,x > 3 \hfill \\ \hfill \\ the\,\,\,integral\,has\,\,the\,\,form\,\,{x^2} - {a^2}\,\, \hfill \\ \hfill \\ x = 3\sec \theta \,\,\,\,\,\, \to \,\,\,dx = 3\sec \theta \tan \theta d\theta \hfill \\ and\,\,\,\sqrt {{x^2} - 9} = 3\tan \theta \hfill \\ \hfill \\ apply\,\,the\,\,substitution \hfill \\ \hfill \\ \int_{}^{} {\frac{{\sqrt {{x^2} - 9} }}{x}} dx\,\, = \int_{}^{} {\frac{{3\tan \theta }}{{3\sec \theta }}\,\left( {3\sec \theta \tan \theta } \right)d\theta } \hfill \\ \hfill \\ = 3\int_{}^{} {{{\tan }^2}\theta d\theta } \, = \,3\int_{}^{} {\,\left( {{{\sec }^2}\theta - 1} \right)d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = 3\,\left( {\tan \theta - \theta } \right) + C \hfill \\ \hfill \\ substitute\,\,for\,\,\tan \theta {\text{ and }}\theta \hfill \\ \hfill \\ = 3\,\left( {\frac{{\sqrt {{x^2} - 9} }}{3}} \right) - 3\operatorname{arcsec} \,\left( {\frac{x}{3}} \right) + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \sqrt {{x^2} - 9} - 3\operatorname{arcsec} \,\left( {\frac{x}{3}} \right) + C \hfill \\ \hfill \\ \end{gathered}$