Answer
$$-\frac{4x^2+1}{2x}+C$$
Work Step by Step
Given
$$ \int \frac{d x}{x^2 \sqrt{4 x^2+1}} $$
Let $$2 x=\tan \theta \Rightarrow x=\frac{1}{2} \tan \theta, \\d x=\frac{1}{2} \sec ^2 \theta d \theta $$
Then
\begin{aligned}
\int \frac{d x}{x^2 \sqrt{4 x^2+1}} &=\int \frac{\frac{1}{2} \sec ^2 \theta d \theta}{\frac{1}{2} \tan^2 \theta \sec \theta}\\
&=\int \frac{\sec \theta}{\tan^2 \theta} d \theta\\
&=\int \frac{1}{\cos \theta}\frac{\cos^2\theta}{\sin^2\theta}d\theta\\
&= \int \sin^{-2}\theta \cos \theta d\theta\\
&= -(\sin \theta)^{-1}+C\\
&= \frac{-1}{\sin \theta}+C\\
&= -\csc \theta +C \\
&= -\frac{4x^2+1}{2x}+C
\end{aligned}
