Answer
$$\int\frac{\sqrt{x^2-4}}{x}dx=\sqrt{x^2-4}-2\sec^{-1}\Big|\frac{x}{2}\Big|+C$$
Work Step by Step
$$A=\int\frac{\sqrt{x^2-4}}{x}dx$$
Use Formula 58, which states that $$\int \frac{\sqrt{x^2-a^2}}{x}dx=\sqrt{x^2-a^2}-a\sec^{-1}\Big|\frac{x}{a}\Big|+C$$
for $a=2$ here:
$$A=\sqrt{x^2-4}-2\sec^{-1}\Big|\frac{x}{2}\Big|+C$$
