Answer
$$\frac{1}{5}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{5}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} - 6x + 34}}} \cr
& {\text{completing the square}} \cr
& = \int {\frac{{dx}}{{{x^2} - 6x + 9 + 25}}} = \int {\frac{{dx}}{{\left( {{x^2} - 6x + 9} \right) + 25}}} \cr
& = \int {\frac{{dx}}{{{{\left( {x - 3} \right)}^2} + 25}}} \cr
& u = x - 3,{\text{ }}du = dx \cr
& = \int {\frac{{du}}{{{u^2} + 25}}} \cr
& {\text{substitute }}u = 5\tan \theta ,{\text{ }}du = 5{\sec ^2}\theta d\theta \cr
& \int {\frac{{du}}{{{u^2} + 25}}} = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{{{\left( {5\tan \theta } \right)}^2} + 25}}} \cr
& = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{25{{\tan }^2}\theta + 25}}} = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{25\left( {{{\tan }^2}\theta + 1} \right)}}} \cr
& {\text{pythagorean identity}} \cr
& = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{25\left( {{{\sec }^2}\theta } \right)}}} = \frac{1}{5}\int {d\theta } \cr
& {\text{integrating}} \cr
& = \frac{1}{5}\theta + C \cr
& u = 5\tan \theta \cr
& = \frac{1}{5}{\tan ^{ - 1}}\left( {\frac{u}{5}} \right) + C \cr
& u = x - 3 \cr
& = \frac{1}{5}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{5}} \right) + C \cr} $$
