Answer
$\ln |\dfrac{\sqrt{9+x^2}}{3}+\dfrac{x}{3}|+C$
Work Step by Step
Let us consider $x =3 \tan \theta $
and $dx= 3 \sec^2 \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{3 \sec^2 \theta d \theta }{\sqrt {9+9 \tan^2 \theta}} =\int \dfrac{3 \sec^2 \theta d \theta }{\sqrt {9 \sec^2 \theta}}$
or, $=\int \sec \theta d \theta $
Now, integrate.
we have $\int \sec \theta d \theta=\ln |\sec \theta +\tan \theta|+C$
Plug $\tan \theta =\dfrac{x}{3}$
Thus, we have
$\ln |\sec \theta +\tan \theta|+C=\ln |\dfrac{\sqrt{9+x^2}}{3}+\dfrac{x}{3}|+C$
