Answer
See the proof below.
Work Step by Step
Since $u=x/3$, then $du= \frac{1}{3}dx$, and hence we get
$$\int \frac{ dx}{9+x^2}=\int \frac{3 du}{9+9u^2}=\frac{1}{3}\int \frac{ du}{1+u^2}\\
=\frac{1}{3} \tan^{-1}u+c=\frac{1}{3} \tan^{-1}\frac{x}{3}+c.$$
