Answer
$$ - \frac{{\sqrt {9 - {w^2}} }}{w} - {\sin ^{ - 1}}\left( {\frac{w}{3}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {9 - {w^2}} }}{{{w^2}}}} dw \cr
& {\text{Using Trigonometric Substitutions}} \cr
& {\text{We set}}{\text{, }}w = 3\sin \theta,{\text{ }}dw = 3\cos \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{\sqrt {9 - {w^2}} }}{{{w^2}}}} dw = \int {\frac{{\sqrt {9 - {{\left( {3\sin \theta } \right)}^2}} }}{{{{\left( {3\sin \theta } \right)}^2}}}} \left( {3\cos \theta d\theta } \right) \cr
& = \int {\frac{{\sqrt {9 - 9{{\sin }^2}\theta } }}{{9{{\sin }^2}\theta }}} \left( {3\cos \theta d\theta } \right) \cr
& = \int {\frac{{\sqrt {9\left( {1 - {{\sin }^2}\theta } \right)} }}{{9{{\sin }^2}\theta }}} \left( {3\cos \theta d\theta } \right) \cr
& = \int {\frac{{3\sqrt {1 - {{\sin }^2}\theta } }}{{9{{\sin }^2}\theta }}} \left( {3\cos \theta d\theta } \right) \cr
& {\text{use the fundamental identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr
& = \int {\frac{{3\sqrt {{{\cos }^2}\theta } }}{{9{{\sin }^2}\theta }}} \left( {3\cos \theta d\theta } \right) \cr
& {\text{simplifying}} \cr
& = \int {\frac{{3\cos \theta }}{{9{{\sin }^2}\theta }}} \left( {3\cos \theta d\theta } \right) \cr
& = \int {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}d\theta } \cr
& {\text{Integrate}} \cr
& = \int {{{\cot }^2}\theta d\theta } \cr
& {\text{use the fundamental identity co}}{{\text{t}}^2}\theta = {\csc ^2}\theta - 1 \cr
& = \int {\left( {{{\csc }^2}\theta - 1} \right)} d\theta \cr
& {\text{Integrate}} \cr
& = - \cot \theta - \theta + C \cr
& {\text{write in terms of }}x,{\text{ }}\cot \theta = \frac{{\sqrt {9 - {w^2}} }}{w}{\text{ and }}\theta = {\sin ^{ - 1}}\left( {\frac{w}{3}} \right) \cr
& = - \frac{{\sqrt {9 - {w^2}} }}{w} - {\sin ^{ - 1}}\left( {\frac{w}{3}} \right) + C \cr} $$
