Answer
$$\displaystyle{\int\frac{\sqrt{x^2-4}}{x}dx=\sqrt{x^2-4}-2\cos^{-1}\left(\frac{2}{x}\right)+C}$$
Work Step by Step
$\displaystyle{I=\int\frac{\sqrt{x^2-4}}{x}dx}$
$\displaystyle \left[\begin{array}{ll} x=2\sec\theta & x^2=4\sec^2\theta \\ & \\ \frac{dx}{d\theta}=2\sec\theta\tan\theta & dx=2\sec\theta\tan\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int\frac{\sqrt{4\sec^2\theta-4}}{2\sec\theta}2\sec\theta\tan\theta\ d\theta}\\
\displaystyle{I=\int\frac{2\tan\theta\times2\sec\theta\tan\theta}{2\sec\theta}d\theta}\\
\displaystyle{I=\int2\tan^2\theta\ d\theta}\\
\displaystyle{I=2\int\left(\sec^2\theta-1\right)\ d\theta}\\
\displaystyle{I=2\left(\tan\theta-\theta\right)+C}\\
\displaystyle{I=2\tan\theta-2\theta+C}$
$\tan\theta=\frac{\sqrt{x^2-4}}{2}\\
\theta=cos^{-1}\left(\frac{2}{x}\right)$
$\displaystyle{I=\sqrt{x^2-4}-2\cos^{-1}\left(\frac{2}{x}\right)+C}$
