Answer
$\ln |\sqrt{9+x^2}+3x|+C$
Work Step by Step
Let us consider $x =\dfrac{1}{3} \tan \theta $
and $dx= \dfrac{1}{3} \sec^2 \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{3 \sec^2 \theta d \theta }{\sqrt {9+9(\dfrac{1}{9} \tan^2 \theta)}} =\int \dfrac{\sec^2 \theta d \theta }{\sqrt { \sec^2 \theta}}$
or, $=\int \sec \theta d \theta $
Now, integrate.
we have $\int \sec \theta d \theta=\ln |\sec \theta +\tan \theta|+C$
Plug in $\tan \theta =3x$
Thus, we have
$\ln |\sec \theta +\tan \theta|+C=\ln |\sqrt{9+x^2}+3x|+C$
