Answer
$$-\ln \left|\frac{\sqrt{4 x^2+1}}{2 x}+\frac{1}{2 x}\right|+C$$
Work Step by Step
Given
$$\int \frac{d x}{x \sqrt{4 x^2+1}} $$
Using the trigeometric subistiution
$$2 x=\tan \theta \Rightarrow x=\frac{1}{2} \tan \theta,\\ d x=\frac{1}{2} \sec ^2 \theta d \theta $$
Then
\begin{aligned}
\int \frac{d x}{x \sqrt{4 x^2+1}} &=\int \frac{\frac{1}{2} \sec ^2 \theta d \theta}{\frac{1}{2} \tan \theta \sec \theta}\\
&=\int \frac{\sec \theta}{\tan \theta} d \theta\\
&=\int \csc \theta d \theta \\
&=-\ln |\csc \theta+\cot \theta|+C \\
&=-\ln \left|\frac{\sqrt{4 x^2+1}}{2 x}+\frac{1}{2 x}\right|+C
\end{aligned}
