Answer
$$\frac{\pi}{2} $$
Work Step by Step
\begin{align*}
\int_{1}^{\infty} \frac{d x}{x \sqrt{x^{2}-1}}&=\int_{1}^{2} \frac{d x}{x \sqrt{x^{2}-1}}+\int_{2}^{\infty} \frac{d x}{x \sqrt{x^{2}-1}}\\
&=\lim _{b \rightarrow 1^{+}} \int_{b}^{2} \frac{d x}{x \sqrt{x^{2}-1}}+\lim _{c \rightarrow \infty} \int_{2}^{c} \frac{d x}{x \sqrt{x^{2}-1}}\\
&=\lim _{b \rightarrow 1^{+}}\left[\sec ^{-1}|x|\right]_{b}^{2}+\left.\lim _{c \rightarrow \infty}\left[\sec ^{-1} | x\right]\right|_{2} ^{c}\\
&=\lim _{b \rightarrow 1^{+}}\left(\sec ^{-1} 2-\sec ^{-1} b\right)+\lim _{c \rightarrow \infty}\left(\sec ^{-1} c-\sec ^{-1} 2\right)\\
&=\left(\frac{\pi}{3}-0\right)+\left(\frac{\pi}{2}-\frac{\pi}{3}\right)\\
&=\frac{\pi}{2}
\end{align*}
