Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_2^\infty {\frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}} \cr
& {\text{By the definition of the improper integrals, we have }} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx \cr
& {\text{then}}{\text{,}} \cr
& \int_2^\infty {\frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}} = \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}} \cr
& \cr
& \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}} = \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{1}{{x - 1}}} \right)_2^b \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{{b - 1}} - \frac{1}{{2 - 1}}} \right) \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{{b - 1}} - 1} \right) \cr
& \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& = - \left( {\frac{1}{{\infty - 1}} - 1} \right) \cr
& = - \left( {0 - 1} \right) \cr
& = 1 \cr} $$
