Answer
$$\ln \left|\sqrt{1+9x^{2}}+3x\right|+c$$
Work Step by Step
Given $$
\int \frac{3d x}{\sqrt{1+9x^{2}}}
$$
Let $ 3x= \tan \theta \ \ \Rightarrow \ \ 3dx= \sec^2 \theta $, then
\begin{align*}
\int \frac{3d x}{\sqrt{1+9x^{2}}}&=\int \frac{ \sec^2 \theta d\theta }{\sqrt{1+\tan^{2}\theta}}\\
&=\int \frac{ \sec^2 \theta d\theta }{\sqrt{\sec^{2}\theta}}\\
&=\int \sec \theta d\theta \\
&=\ln |\sec \theta +\tan \theta |+ c\\
&=\ln \left|\sqrt{1+9x^{2}}+3x\right|+c
\end{align*}
