## Calculus (3rd Edition)

Since $u=x/3$, then $du= \frac{1}{3}dx$, and hence we get $$\int \frac{ dx}{9+x^2}=\int \frac{3 du}{9+9u^2}=\frac{1}{3}\int \frac{ du}{1+u^2}\\ =\frac{1}{3} \tan^{-1}u+c=\frac{1}{3} \tan^{-1}\frac{x}{3}+c.$$