Answer
$$\frac{1}{6}\ln \left| {\frac{{3 + x}}{{3 - x}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{9 - {x^2}}}} \cr
& \left( {\bf{a}} \right){\text{using the partial fractions method}} \cr
& \,\,\,\int {\frac{{dx}}{{9 - {x^2}}}} = \int {\frac{{dx}}{{\left( {3 + x} \right)\left( {3 - x} \right)}}} \cr
& \,\,\frac{1}{{\left( {3 + x} \right)\left( {3 - x} \right)}} = \frac{A}{{3 + x}} + \frac{B}{{3 - x}} \cr
& \,\,\,{\text{multiply by }}\left( {3 + x} \right)\left( {3 - x} \right) \cr
& \,\,\,\,\,1 = A\left( {3 - x} \right) + B\left( {3 + x} \right) \cr
& \,\,\,\,{\text{If }}x = - 3,\,\,\,\,1 = A\left( 6 \right),\,\,\,\,A = 1/6 \cr
& \,\,\,\,{\text{If }}x = 3,\,\,\,\,1 = B\left( 6 \right),\,\,\,\,B = 1/6 \cr
& \cr
& \,\,\frac{1}{{\left( {3 + x} \right)\left( {3 - x} \right)}} = \frac{{1/6}}{{3 + x}} + \frac{{1/6}}{{3 - x}} \cr
& \,\,Then, \cr
& \int {\frac{{dx}}{{9 - {x^2}}}} = \int {\left( {\frac{{1/6}}{{3 + x}} + \frac{{1/6}}{{3 - x}}} \right)} dx \cr
& {\text{integrate }} \cr
& = \frac{1}{6}\ln \left| {3 + x} \right| - \frac{1}{6}\ln \left| {3 - x} \right| + C \cr
& {\text{Factoring}} \cr
& = \frac{1}{6}\left( {\ln \left| {3 + x} \right| - \ln \left| {3 - x} \right|} \right) + C \cr
& = \frac{1}{6}\ln \left| {\frac{{3 + x}}{{3 - x}}} \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}}{{9 - {x^2}}}} = \int {\frac{{3\cos \theta d\theta }}{{9 - {{\left( {3\sin \theta } \right)}^2}}}} \cr
& \,\,\,\, = \int {\frac{{3\cos \theta d\theta }}{{9 - 9{{\sin }^2}\theta }}} \cr
& \,\,\,\, = \frac{1}{3}\int {\frac{{\cos \theta d\theta }}{{{{\cos }^2}\theta }}} \cr
& \,\,\,\, = \frac{1}{3}\int {\sec \theta } d\theta \cr
& \,\,\, = \frac{1}{3}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& \,\,{\text{Where sec}}\theta = \frac{3}{{\sqrt {9 - {x^2}} }}{\text{ and tan}}\theta = \frac{x}{{\sqrt {9 - {x^2}} }} \cr
& \,\,\, = \frac{1}{3}\ln \left| {\frac{3}{{\sqrt {9 - {x^2}} }} + \frac{x}{{\sqrt {9 - {x^2}} }}} \right| + C \cr
& \,\,\, = \frac{1}{3}\ln \left| {\frac{{3 + x}}{{\sqrt {9 - {x^2}} }}} \right| + C \cr
& \,\,\,\,\,{\text{simplifying}} \cr
& \,\,\, = \frac{1}{3}\ln \left| {\frac{{3 + x}}{{\sqrt {3 + x} \sqrt {3 - x} }}} \right| + C \cr
& \,\,\, = \frac{1}{3}\ln \left| {\frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }}} \right| + C \cr
& \,\,\, = \frac{1}{3}\left( {\frac{1}{2}} \right)\ln \left| {\frac{{3 + x}}{{3 - x}}} \right| + C \cr
& \,\,\, = \frac{1}{6}\ln \left| {\frac{{3 + x}}{{3 - x}}} \right| + C \cr} $$
