Answer
$$\frac{1}{2} \sec ^{-1}\left(\frac{x}{2}\right)+C$$
Work Step by Step
Given $$\int \frac{d x}{x \sqrt{x^{2}-4}}$$
Let
$$x=2\sec u \ \ \ \ \ \ \ \ dx=2\sec u\tan udu $$
Then
\begin{align*}
\int \frac{d x}{x \sqrt{x^{2}-4}}&=\int \frac{2\sec u\tan udu}{2\sec u \sqrt{4\sec^2 u-4}}\\
&= \int \frac{2\sec u\tan udu}{2\sec u \sqrt{4\tan^2 u}}\\
&=\frac{1}{2}\int du\\
&=\frac{1}{2}u+C\\
&= \frac{1}{2} \sec ^{-1}\left(\frac{x}{2}\right)+C
\end{align*}
