Answer
$${\tan ^{ - 1}}\left( {y - 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dy}}{{{y^2} - 2y + 2}}} \cr
& {\text{Complete the square in the denominator}} \cr
& {y^2} - 2y + 2 = {y^2} - 2y + 1 + 1 \cr
& {y^2} - 2y + 2 = {\left( {y - 1} \right)^2} + 1 \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{dy}}{{{y^2} - 2y + 2}}} = \int {\frac{{dy}}{{{{\left( {y - 1} \right)}^2} + 1}}} \cr
& {\text{Let }}u = y - 1,\,\,\,du = dy \cr
& = \int {\frac{{du}}{{{u^2} + 1}}} \cr
& \cr
& {\text{Integrating}} \cr
& \int {\frac{{du}}{{{u^2} + 1}}} = {\tan ^{ - 1}}u + C \cr
& {\text{Write in terms of }}y{\text{; replace }}y - 1{\text{ for }}u \cr
& = {\tan ^{ - 1}}\left( {y - 1} \right) + C \cr} $$
