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