Calculus: Early Transcendentals (2nd Edition)

$= \frac{{\sqrt {9{x^2} - 1} }}{x} + C$
$\begin{gathered} \int_{}^{} {\frac{{dx}}{{{x^2}\sqrt {9{x^2} - 1} }}} \hfill \\ \hfill \\ the\,integral\,contains\,\,the\,\,form\,{\text{ }}{u^2} - {x^2}\,\, \hfill \\ then \hfill \\ set\,\,3x = \sec \theta \,\,\,\,\, \to \,\,\,\,x = \frac{1}{3}\sec \theta \hfill \\ and\,\,\sqrt {9{x^2} - 1} = \tan \theta \hfill \\ \hfill \\ Substituting{\text{ }}these{\text{ }}factors{\text{ }} \hfill \\ \hfill \\ = \int_{}^{} {\frac{{\frac{1}{3}\sec \theta \tan \theta d\theta }}{{\,{{\left( {\frac{1}{3}\sec \theta } \right)}^2}\,\left( {\tan \theta } \right)}}} = \int_{}^{} {\frac{1}{{\sec \theta }}d\theta } \hfill \\ \hfill \\ = 3\int_{}^{} {\cos \theta d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = 3\sin \theta + C \hfill \\ \hfill \\ substitute\,\,for\,\,\sin \theta {\text{ }} \hfill \\ \hfill \\ \sin \theta = \frac{{\sqrt {9{x^2} - 1} }}{{3x}} \hfill \\ \hfill \\ = 3\,\left( {\frac{{\sqrt {9{x^2} - 1} }}{{3x}}} \right) + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{\sqrt {9{x^2} - 1} }}{x} + C \hfill \\ \hfill \\ \end{gathered}$