Answer
$\displaystyle \sqrt{3}-\frac{\pi}{3}$
Work Step by Step
Trigonometric substitution:
$\displaystyle \sqrt{x^{2}-a^{2}}\quad x=a\sec\theta,\quad 0\leq\theta\lt \frac{\pi}{2}$ or $\displaystyle \pi\leq\theta\lt \frac{3\pi}{2}$
Identity: $\sec^{2}\theta-1=\tan^{2}\theta$
$\displaystyle \int_{1}^{2}\frac{\sqrt{x^{2}-1}}{x}dx=\left[\begin{array}{ll}
x=\sec\theta & \\
dx=\sec\theta\tan\theta d\theta & \\
& \sqrt{x^{2}-1}=\tan\theta\\
x=1\rightarrow\theta=0, & x=2\rightarrow\theta=\pi/3
\end{array}\right]$
$\displaystyle \int_{0}^{\pi/3}\frac{\tan\theta}{\sec\theta}\sec\theta\tan\theta d\theta$
$=\displaystyle \int_{0}^{\pi/3}\tan^{2}\theta d\theta$
... apply the identity
$=\displaystyle \int_{0}^{\pi/3}\left(\sec^{2}\theta-1\right)d\theta$
$=[\tan\theta-\theta]_{0}^{\pi/3}$
$=\displaystyle \sqrt{3}-\frac{\pi}{3}$
