Answer
$${\tan ^{ - 1}}\left( {x - 2} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} - 4x + 5}}} \cr
& {\text{completing the square}} \cr
& \int {\frac{{dx}}{{{x^2} - 4x + 5}}} = \int {\frac{{dx}}{{{x^2} - 4x + 4 + 1}}} \cr
& = \int {\frac{{dx}}{{{{\left( {x - 2} \right)}^2} + 1}}} \cr
& \cr
& {\text{Write in terms of }}\theta \cr
& {\text{substitute }}x = \tan \theta + 2,{\text{ }}\,\,\,dx = {\sec ^2}\theta d\theta \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {\tan \theta + 2 - 2} \right)}^2} + 1}}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\tan }^2}\theta + 1}}} \cr
& = \int {\frac{{{{\sec }^2}\theta }}{{{{\sec }^2}\theta }}} d\theta \cr
& = \int {d\theta } \cr
& {\text{Integrate}} \cr
& = \theta + C \cr
& \cr
& {\text{Where }}x = \tan \theta + 2 \to \theta = {\tan ^{ - 1}}\left( {x - 2} \right) \cr
& = {\tan ^{ - 1}}\left( {x - 2} \right) + C \cr} $$
