Chapter 8: Techniques of Integration - Section 8.8 - Improper Integrals - Exercises 8.8 - Page 501: 29

$$\frac{\pi}{3}$$

\begin{align*}\int_{1}^{2} \frac{d s}{_{s \sqrt{s^{2}-1}}}&=\lim _{b \rightarrow 1^{+}}\int_{b}^{2} \frac{d s}{_{s \sqrt{s^{2}-1}}}\\ &=\lim _{b \rightarrow 1^{+}}\left[\sec ^{-1} s\right]\bigg|_{b}^{2}\\ &=\lim _{b \rightarrow 1^{+}}\left[\sec ^{-1} 2-\sec ^{-1} b\right]\\ &=\frac{\pi}{3}-0=\frac{\pi}{3} \end{align*}