Answer
$\ln |\dfrac{\sqrt{9+x^2}}{3}+\dfrac{x}{3}|+C$
Work Step by Step
Plug $x =3 \tan \theta \implies dx= 3 \sec^2 \theta d \theta $
Then,
$\int \dfrac{3 \sec^2 \theta d \theta }{\sqrt {9+9 \tan^2 \theta}} =\int \dfrac{3 \sec^2 \theta d \theta }{\sqrt {9 \sec^2 \theta}}$
or, $\int \sec \theta d \theta=\ln |\sec \theta +\tan \theta|+C$
We have $ x =3 \tan \theta \implies \tan \theta =(\dfrac{x}{3})$
Thus,
$\ln |\sec \theta +\tan \theta|+C=\ln |\dfrac{\sqrt{9+x^2}}{3}+(\dfrac{x}{3})|+C$
