Answer
$$\frac{1}{2}\ln \left( {{t^2} - 6t + 10} \right) + {\tan ^{ - 1}}\left( {t - 3} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{t - 2}}{{{t^2} - 6t + 10}}} dt \cr
& {\text{complete the square for }}{t^2} - 6t + 10 \cr
& = \int {\frac{{t - 2}}{{\left( {{t^2} - 6t + 9} \right) + 1}}} dt \cr
& = \int {\frac{{t - 2}}{{{{\left( {t - 3} \right)}^2} + 1}}} dt \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = t - 3,{\text{ so that }}du = dt \cr
& \int {\frac{{t - 2}}{{{{\left( {t - 3} \right)}^2} + 1}}} dt = \int {\frac{{u + 3 - 2}}{{{u^2} + 1}}} du \cr
& = \int {\frac{{u + 1}}{{{u^2} + 1}}} du \cr
& = \int {\frac{u}{{{u^2} + 1}}} du + \int {\frac{1}{{{u^2} + 1}}} du \cr
& {\text{integrate by using the formulas}} \cr
& \int {\frac{{du}}{u}} = \ln \left| u \right| + C{\text{ and }}\int {\frac{1}{{{u^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + C \cr
& {\text{then}} \cr
& = \frac{1}{2}\ln \left( {{u^2} + 1} \right) + {\tan ^{ - 1}}u + C \cr
& {\text{write in terms of }}t;{\text{ replace }}t - 3{\text{ for }}u \cr
& = \frac{1}{2}\ln \left( {{t^2} - 6t + 10} \right) + {\tan ^{ - 1}}\left( {t - 3} \right) + C \cr} $$
