Answer
$$2\pi $$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{{8dx}}{{{x^2} - 2x + 2}}} \cr
& {\text{complete the square for }}{x^2} - 2x + 2 \cr
& = {x^2} - 2x + 1 + 1 \cr
& = \left( {{x^2} - 2x + 1} \right) + 1 \cr
& = {\left( {x - 1} \right)^2} + 1 \cr
& = \int_1^2 {\frac{{8dx}}{{{{\left( {x - 1} \right)}^2} + 1}}} \cr
& {\text{use the substitution method}}{\text{:}} \cr
& u = x - 1,{\text{ so that }}du = dx \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 2,{\text{ then }}u = 2 - 1 = 1 \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = 1 - 1 = 0 \cr
& {\text{then}} \cr
& \int_1^2 {\frac{{8dx}}{{{{\left( {x - 1} \right)}^2} + 1}}} = \int_0^1 {\frac{{8du}}{{{u^2} + 1}}} \cr
& = 8\int_0^1 {\frac{{du}}{{{u^2} + 1}}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{{u^2} + {a^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = 8\left( {{{\tan }^{ - 1}}u} \right)_0^1 \cr
& = 8\left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right) \cr
& = 8\left( {\frac{\pi }{4} - 0} \right) \cr
& = 2\pi \cr} $$
