Answer
$$ \frac{\sqrt{4x^2-1}}{x}+c$$
Work Step by Step
Given
$$\int \frac{d x}{x^2 \sqrt{4 x^2-1}}$$
AApply Integral Substitution
\begin{aligned}
2x&= \sec u \\
2dx&=\sec u\tan udu
\end{aligned}
Then
\begin{aligned}
\int \frac{d x}{x^2 \sqrt{4 x^2-1}}&=\frac{4}{2}\int \frac{\sec u\tan udu}{\sec^2u \sqrt{\sec^2u-1}}\\
&= 2\int \frac{du}{\sec u}\\
&= 2\int \cos udu\\
&= 2\sin (u)+c\\
&= \frac{2\sqrt{4x^2-1}}{2x}+c
\end{aligned}
