Answer
$(\dfrac{1}{3}) \ln |\dfrac{3+x}{\sqrt {9-x^2}}|+c$
or:
$(\dfrac{1}{6}) \ln |\dfrac{x+3}{x-3}|+c$
Work Step by Step
Consider the integral $\int \dfrac{dx}{(9-x^2)}$
The given integral can be re-written as: $\int \dfrac{dx}{(9-x^2)}=\int \dfrac{dx}{(\sqrt {9-x^2)^2}}$
Plug in $x= 3 \sin \theta \implies dx= 3 \cos \theta d \theta$ and $\sqrt {9-x^2}=3 \cos \theta$
This implies that $\int\dfrac{3 \cos \theta}{(3 \cos \theta)^2}=(\dfrac{1}{3}) \int \sec \theta d\theta$
$(\dfrac{1}{3}) \ln |\sec \theta+\tan \theta|+c=(\dfrac{1}{3}) \ln |\dfrac{3}{\sqrt {9-x^2}}+\dfrac{x}{\sqrt {9-x^2}}|+c$
or: $(\dfrac{1}{3}) \ln |\dfrac{3+x}{\sqrt {9-x^2}}|+c$
or: $(\dfrac{1}{6}) \ln |\dfrac{x+3}{x-3}|+c$
