## Thomas' Calculus 13th Edition

$$1$$
\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}