Calculus (3rd Edition)

$$\frac{1}{3}\sinh^{-1}3x+c.$$
We have $$\int\frac{1}{\sqrt{1+9x^2}}dx=\int\frac{1}{\sqrt{1+(3x)^2}}dx.$$ Let $u=3x$, then $du=3dx$ and since $\frac{d}{dx} \sinh^{-1}x=\frac{1}{\sqrt{x^2+1}}$, we obtain $$\int\frac{1}{\sqrt{1+(3x)^2}}dx=\frac{1}{3} \int\frac{1}{\sqrt{1+u^2}}du= \frac{1}{3}\sinh^{-1}u+c\\ = \frac{1}{3}\sinh^{-1}3x+c.$$