Answer
$\sqrt{x^2-4}-2\sec^{-1}|\dfrac{x}{2}\Big|+C$
Work Step by Step
Apply formula : $\int \dfrac{\sqrt{x^2-a^2}}{x}dx=\sqrt{x^2-a^2}-a\sec^{-1} |\dfrac{x}{a}|+C$
Consider $I=\int\dfrac{\sqrt{x^2-4}}{x}dx \\=\sqrt{x^2-4}-2\sec^{-1}|\dfrac{x}{2}|+C$