Answer
$\dfrac{\pi}{3}$
Work Step by Step
Here, we have $\int_{1}^{2} \dfrac{ds}{s \sqrt {s^2-1}}=\lim\limits_{a \to 1^{+}}\int_{a}^{2} \dfrac{ds}{s \sqrt {s^2-1}}$
This implies that
$\lim\limits_{a \to 1^{+}}\int_{a}^{2} \dfrac{ds}{s \sqrt {s^2-1}}=\lim\limits_{a \to 1^{+}}[\sec^{-1} s]_{a}^{2}$
Thus, $\lim\limits_{a \to 1^{+}}\sec^{-1} (2)-\lim\limits_{a \to 1^{+}}\sec^{-1} (a)=\dfrac{\pi}{3}$
