Answer
$$2$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{{dx}}{{\sqrt {x - 1} }}} \cr
& {\text{The integrand is not defined for }}x = 1,{\text{ then}} \cr
& \int_1^2 {\frac{{dx}}{{\sqrt {x - 1} }}} = \mathop {\lim }\limits_{a \to {1^ + }} \int_a^2 {\frac{{dx}}{{\sqrt {x - 1} }}} \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{a \to {1^ + }} \left[ {2\sqrt {x - 1} } \right]_a^2 \cr
& = \mathop {\lim }\limits_{a \to {1^ + }} \left[ {2\sqrt {2 - 1} - 2\sqrt {a - 1} } \right] \cr
& = \mathop {\lim }\limits_{a \to {1^ + }} \left[ {2 - 2\sqrt {a - 1} } \right] \cr
& {\text{Evaluating the limit}} \cr
& = 2 - 2\sqrt {\left( 1 \right) - 1} \cr
& = 2 \cr} $$
