Answer
$$\frac{1}{3}\left[ {{{\tan }^{ - 1}}\left( {\frac{5}{3}} \right) - {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\frac{{dt}}{{{t^2} - 2t + 10}}} \cr
& {\text{Completing the square}} \cr
& = \int {\frac{{dt}}{{{t^2} - 2t + 1 + 9}}} \cr
& = \int {\frac{{dt}}{{{{\left( {t + 1} \right)}^2} + {3^2}}}} \cr
& {\text{The integrand contains the form }}{u^2} + {a^2} \cr
& {\text{Use the change of variable }}t + 1 = 3\tan \theta \cr
& t + 1 = 3\tan \theta ,\,\,\,dt = 3{\sec ^2}\theta d\theta \cr
& {\text{Use the change of variable}} \cr
& = \int {\frac{{3{{\sec }^2}\theta d\theta }}{{{{\left( {3\tan \theta } \right)}^2} + {3^2}}}} \cr
& = \int {\frac{{3{{\sec }^2}\theta d\theta }}{{9{{\tan }^2}\theta + 9}}} \cr
& = \frac{3}{9}\int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\tan }^2}\theta + 1}}} \cr
& = \frac{1}{3}\int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\sec }^2}\theta }}} \cr
& = \frac{1}{3}\int {d\theta } \cr
& {\text{Integrating}} \cr
& = \frac{1}{3}\theta + C \cr
& {\text{Write in terms of }}t \cr
& = \frac{1}{3}{\tan ^{ - 1}}\left( {\frac{{t + 1}}{3}} \right) + C \cr
& \cr
& ,{\text{then}} \cr
& \int_1^4 {\frac{{dt}}{{{t^2} - 2t + 10}}} = \left[ {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{{t + 1}}{3}} \right)} \right]_1^4 \cr
& = \left[ {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{{4 + 1}}{3}} \right)} \right] - \left[ {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{{1 + 1}}{3}} \right)} \right] \cr
& = \frac{1}{3}{\tan ^{ - 1}}\left( {\frac{5}{3}} \right) - \frac{1}{3}{\tan ^{ - 1}}\left( {\frac{2}{3}} \right) \cr
& = \frac{1}{3}\left[ {{{\tan }^{ - 1}}\left( {\frac{5}{3}} \right) - {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right] \cr} $$
