Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{{dx}}{{{x^2} - 6x + 9}}} \cr
& {\text{perfect square trinomial }}{\left( {x - 3} \right)^2} = {x^2} - 6x + 9 \cr
& \int_1^2 {\frac{{dx}}{{{x^2} - 6x + 9}}} = \int_1^2 {\frac{{dx}}{{{{\left( {x - 3} \right)}^2}}}} \cr
& u = x - 3,{\text{ then }}du = dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 2,{\text{ }}u = 2 - 3 = - 1 \cr
& {\text{if }}x = 1,{\text{ }}u = 1 - 3 = - 2 \cr
& {\text{so}} \cr
& \int_1^2 {\frac{{dx}}{{{{\left( {x - 3} \right)}^2}}}} = \int_{ - 2}^{ - 1} {\frac{{du}}{{{u^2}}}} \cr
& = \int_{ - 2}^{ - 1} {{u^{ - 2}}du} \cr
& {\text{find the antiderivative }} \cr
& = \left( {\frac{{{u^{ - 1}}}}{{ - 1}}} \right)_{ - 2}^{ - 1} \cr
& = - \left( {\frac{1}{u}} \right)_{ - 2}^{ - 1} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \frac{1}{{ - 1}} + \frac{1}{{ - 2}} \cr
& = \frac{1}{2} \cr} $$
