Answer
$$\frac{1}{9}\ln \left| {\frac{x}{{\sqrt {9 - {x^2}} }}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\left( {9 - {x^2}} \right)}}} \cr
& \left( {\bf{a}} \right){\text{using the partial fractions method}} \cr
& \,\,\,\int {\frac{{dx}}{{x\left( {9 - {x^2}} \right)}}} = \int {\frac{{dx}}{{x\left( {3 + x} \right)\left( {3 - x} \right)}}} \cr
& \,\,\frac{1}{{x\left( {3 + x} \right)\left( {3 - x} \right)}} = \frac{A}{x} + \frac{B}{{3 + x}} + \frac{C}{{3 - x}} \cr
& \,\,\,{\text{multiply by }}x\left( {3 + x} \right)\left( {3 - x} \right) \cr
& \,\,\,\,\,1 = A\left( {3 + x} \right)\left( {3 - x} \right) + Bx\left( {3 - x} \right) + Cx\left( {3 + x} \right) \cr
& \,\,\,\,{\text{If }}x = 0,\,\,\,\,1 = A\left( 9 \right),\,\,\,\,A = 1/9 \cr
& \,\,\,\,{\text{If }}x = - 3,\,\,\,\,1 = B\left( { - 18} \right),\,\,\,\,B = - 1/18 \cr
& \,\,\,\,{\text{If }}x = 3,\,\,\,\,1 = C\left( {18} \right),\,\,\,\,C = 1/18 \cr
& \,\,\frac{1}{{x\left( {3 + x} \right)\left( {3 - x} \right)}} = \frac{{1/9}}{x} + \frac{{ - 1/18}}{{3 + x}} + \frac{{1/18}}{{3 - x}} \cr
& \,\,Then, \cr
& \int {\frac{{dx}}{{x\left( {9 - {x^2}} \right)}}} = \int {\left( {\frac{{1/9}}{x} + \frac{{ - 1/18}}{{3 + x}} + \frac{{1/18}}{{3 - x}}} \right)} dx \cr
& {\text{integrate }} \cr
& = \frac{1}{9}\ln \left| x \right| - \frac{1}{{18}}\ln \left| {3 + x} \right| - \frac{1}{{18}}\ln \left| {3 - x} \right| + C \cr
& = \frac{1}{9}\ln \left| x \right| - \frac{1}{{18}}\ln \left| {9 - {x^2}} \right| + C \cr
& {\text{Factoring}} \cr
& = \frac{1}{9}\left( {\ln \left| x \right| - \frac{1}{2}\ln \left| {9 - {x^2}} \right|} \right) + C \cr
& = \frac{1}{9}\left( {\ln \left| x \right| - \ln \sqrt {9 - {x^2}} } \right) + C \cr
& = \frac{1}{9}\ln \left| {\frac{x}{{\sqrt {9 - {x^2}} }}} \right| + 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{{dx}}{{x\left( {9 - {x^2}} \right)}}} = \int {\frac{{3\cos \theta d\theta }}{{\left( {3\sin \theta } \right)\left( {9 - {{\left( {3\sin \theta } \right)}^2}} \right)}}} \cr
& = \int {\frac{{3\cos \theta d\theta }}{{\left( {3\sin \theta } \right)\left( {9 - 9{{\sin }^2}\theta } \right)}}} \cr
& = \frac{1}{9}\int {\frac{{\cos \theta d\theta }}{{\sin \theta \left( {1 - {{\sin }^2}\theta } \right)}}} \cr
& = \frac{1}{9}\int {\frac{{\cos \theta d\theta }}{{\sin \theta {{\cos }^2}\theta }}} \cr
& = \frac{1}{9}\int {\frac{1}{{\sin \theta \cos \theta }}} d\theta \cr
& = \frac{1}{9}\int {\frac{2}{{2\sin \theta \cos \theta }}} d\theta \cr
& {\text{Use the identity }}2\sin x = 2\sin x\cos x \cr
& = \frac{1}{9}\int {\frac{1}{{\sin 2\theta }}} d\theta \cr
& = \frac{1}{9}\int {\csc 2\theta \left( 2 \right)} d\theta \cr
& {\text{Integrate by the formula }}\int {\csc x} dx = \ln \left| {\csc x - \cot x} \right| + C \cr
& = \frac{1}{9}\ln \left| {\csc 2\theta - \cot 2\theta } \right| + C \cr
& {\text{Using trigonometric identitities}} \cr
& = \frac{1}{9}\ln \left| {\frac{1}{{\sin 2\theta }} - \frac{{\cos 2\theta }}{{\sin 2\theta }}} \right| + C \cr
& = \frac{1}{9}\ln \left| {\frac{1}{{\sin 2\theta }} - \frac{{\cos 2\theta }}{{\sin 2\theta }}} \right| + C \cr
& = \frac{1}{9}\ln \left| {\frac{{1 - \cos 2\theta }}{{\sin 2\theta }}} \right| + C \cr
& = \frac{1}{9}\ln \left| {\frac{{1 - 1 + 2{{\sin }^2}\theta }}{{2\sin \theta \cos \theta }}} \right| + C \cr
& = \frac{1}{9}\ln \left| {\frac{{2{{\sin }^2}\theta }}{{2\sin \theta \cos \theta }}} \right| + C \cr
& = \frac{1}{9}\ln \left| {\frac{{\sin \theta }}{{\cos \theta }}} \right| + C \cr
& = \frac{1}{9}\ln \left| {\tan \theta } \right| + C \cr
& {\text{Where tan}}\theta = \frac{x}{{\sqrt {9 - {x^2}} }} \cr
& = \frac{1}{9}\ln \left| {\frac{x}{{\sqrt {9 - {x^2}} }}} \right| + C \cr} $$
