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