Answer
$$ \frac{1}{3}\sinh^{-1}3x+c.$$
Work Step by Step
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
$$\int\frac{1}{\sqrt{1+(3x)^2}}dx=\frac{1}{3} \int\frac{1}{\sqrt{1+u^2}}du= \frac{1}{3}\sinh^{-1}u+c\\
= \frac{1}{3}\sinh^{-1}3x+c.$$
