Answer
$$\sqrt{3} -\frac{\pi }{3}$$
Work Step by Step
Given $$\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x} d x $$
Let $u^2=x^2-1\ \Rightarrow 2udu=2xdx $, at $x=1\to u=0$, $x=2\to u=\sqrt{3}$, then
\begin{align*}
\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x} d x&=\int_{0}^{\sqrt{3}} \frac{u^2du}{u^2+1} \\
&=\int_{0}^{\sqrt{3}} \left(1-\frac{1}{u^2+1} \right)du\\
&=u-\tan^{-1}u\bigg|_{0}^{\sqrt{3}}\\
&=\sqrt{3} -\frac{\pi }{3}
\end{align*}
