Answer
$$ - \frac{{\sqrt {9 - {x^2}} }}{{9x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2}\sqrt {9 - {x^2}} }}} \cr
& {\text{substitute }}x = 3\sin \theta ,{\text{ }}dx = 3\cos \theta d\theta \cr
& = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {3\sin \theta } \right)}^2}\sqrt {9 - {{\left( {3\sin \theta } \right)}^2}} }}} \cr
& = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {3\sin \theta } \right)}^2}\sqrt {9 - 9{{\sin }^2}\theta } }}} \cr
& = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {3\sin \theta } \right)}^2}\left( 3 \right)\sqrt {1 - {{\sin }^2}\theta } }}} \cr
& = \frac{1}{9}\int {\frac{{\cos \theta d\theta }}{{{{\sin }^2}\theta \sqrt {{{\cos }^2}\theta } }}} \cr
& = \frac{1}{9}\int {\frac{1}{{{{\sin }^2}\theta }}} d\theta \cr
& = \frac{1}{9}\int {{{\csc }^2}\theta } d\theta \cr
& {\text{find antiderivative}} \cr
& = \frac{1}{9}\left( { - \cot \theta } \right) + C \cr
& = - \frac{1}{9}\cot \theta + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{9}\left( {\frac{{\sqrt {9 - {x^2}} }}{x}} \right) + C \cr
& = - \frac{{\sqrt {9 - {x^2}} }}{{9x}} + C \cr} $$
