Answer
$$\ln \frac{1}{{\sqrt {9 - {x^2}} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{xdx}}{{9 - {x^2}}}} \cr
& \cr
& \left( {\bf{a}} \right){\text{using the substitution method}} \cr
& \,\,\,\,{\text{Let }}u = 9 - {x^2},\,\,\,\,\,du = - 2xdx,\,\,\,\,dx = \frac{{du}}{{ - 2x}} \cr
& \int {\frac{{xdx}}{{9 - {x^2}}}} = \int {\frac{x}{u}\left( {\frac{{du}}{{ - 2x}}} \right)} \cr
& = - \frac{1}{2}\int {\frac{1}{u}du} \cr
& {\text{integrate }} \cr
& = - \frac{1}{2}\ln \left| u \right| + C \cr
& {\text{write in terms of }}x,{\text{ substitute 9}} - {x^2}{\text{ for }}u \cr
& = - \frac{1}{2}\ln \left| {9 - {x^2}} \right| + C \cr
& {\text{using logarithmic properties}} \cr
& = \ln {\left( {9 - {x^2}} \right)^{ - 1/2}} + C \cr
& = \ln \frac{1}{{\sqrt {9 - {x^2}} }} + C \cr
& \cr
& \left( {\bf{b}} \right){\text{using a trigonometric substitution}} \cr
& \,\,\,\,\,{\text{Let }}x = 3\sin \theta ,\,\,\,dx = 3\cos \theta d\theta \cr
& \,\,\,\,\,\int {\frac{{xdx}}{{9 - {x^2}}}} = \int {\frac{{\left( {3\sin \theta } \right)\left( {3\cos \theta d\theta } \right)}}{{9 - {{\left( {3\sin \theta } \right)}^2}}}} \cr
& = \int {\frac{{9\sin \theta \cos \theta d\theta }}{{9 - 9{{\sin }^2}\theta }}} \cr
& = \int {\frac{{\sin \theta \cos \theta d\theta }}{{1 - {{\sin }^2}\theta }}} \cr
& = \int {\frac{{\sin \theta \cos \theta d\theta }}{{{{\cos }^2}\theta }}} \cr
& = \int {\frac{{\sin \theta d\theta }}{{\cos \theta }}} \cr
& = \int {\tan \theta } d\theta \cr
& {\text{Integrate}} \cr
& = - \ln \left| {\cos \theta } \right| + C \cr
& {\text{where }}\cos \theta = \frac{{\sqrt {9 - {x^2}} }}{2} \cr
& = - \ln \left| {\frac{{\sqrt {9 - {x^2}} }}{2}} \right| + C \cr
& {\text{simplify using logaritmic properties and combine the constants}} \cr
& = - \ln \left| {\sqrt {4 - {x^2}} } \right| + \ln \left| 2 \right| + C \cr
& = - \frac{1}{2}\ln \left| {4 - {x^2}} \right| + C \cr
& {\text{using logarithmic properties}} \cr
& = \ln {\left( {9 - {x^2}} \right)^{ - 1/2}} + C \cr
& = \ln \frac{1}{{\sqrt {9 - {x^2}} }} + C \cr} $$
