Answer
$$\frac{1}{2}\ln \left( {{x^2} + 6x + 13} \right) - \frac{5}{2}{\tan ^{ - 1}}\left( {\frac{{x + 3}}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{x - 2}}{{{x^2} + 6x + 13}}dx} \cr
& {\text{Completing the square}} \cr
& = \int {\frac{{x - 2}}{{{x^2} + 6x + 9 + 4}}dx} \cr
& = \int {\frac{{x - 2}}{{{{\left( {x + 3} \right)}^2} + {2^2}}}dx} \cr
& {\text{Let }}u = x + 3,\,\,\,du = dx \cr
& = \int {\frac{{u - 5}}{{{u^2} + {2^2}}}du} \cr
& = \int {\frac{u}{{{u^2} + {2^2}}}du} - \int {\frac{5}{{{u^2} + {2^2}}}du} \cr
& = \frac{1}{2}\int {\frac{{2u}}{{{u^2} + {2^2}}}du} - 5\int {\frac{1}{{{u^2} + {2^2}}}du} \cr
& {\text{Integrating}} \cr
& {\text{ = }}\frac{1}{2}\ln \left( {{u^2} + {2^2}} \right) - 5\left( {\frac{1}{2}} \right){\tan ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr
& {\text{ = }}\frac{1}{2}\ln \left( {{u^2} + 4} \right) - \frac{5}{2}{\tan ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr
& {\text{substitute }}u = x + 3 \cr
& {\text{ = }}\frac{1}{2}\ln \left( {{x^2} + 6x + 13} \right) - \frac{5}{2}{\tan ^{ - 1}}\left( {\frac{{x + 3}}{2}} \right) + C \cr} $$
