Answer
$$\ln \left( {{x^2} + 6x + 13} \right) - 3{\tan ^{ - 1}}\left( {\frac{{x + 3}}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2x}}{{{x^2} + 6x + 13}}} dx \cr
& {\text{Completing the square}} \cr
& \int {\frac{{2x}}{{{x^2} + 6x + 13}}} dx = \int {\frac{{2x}}{{{x^2} + 6x + 9 + 4}}} dx \cr
& = \int {\frac{{2x}}{{{{\left( {x + 3} \right)}^2} + {2^2}}}} dx \cr
& {\text{Let }}u = x + 3,{\text{ }}x = u - 3,{\text{ }}dx = du,{\text{ }} \cr
& {\text{Substituting}} \cr
& = \int {\frac{{2x}}{{{{\left( {x + 3} \right)}^2} + {2^2}}}} dx = \int {\frac{{2\left( {u - 3} \right)}}{{{u^2} + {2^2}}}} dx = \int {\frac{{2u - 6}}{{{u^2} + {2^2}}}} dx \cr
& = \int {\frac{{2u}}{{{u^2} + 4}}} dx - \int {\frac{6}{{{u^2} + 4}}} dx \cr
& {\text{Integrate using basic integration rules}} \cr
& = \ln \left( {{u^2} + 4} \right) - 6\left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{u}{2}} \right) + C \cr
& = \ln \left( {{u^2} + 4} \right) - 3{\tan ^{ - 1}}\frac{u}{2} + C \cr
& {\text{Write in terms of }}x \cr
& = \ln \left( {{{\left( {x + 3} \right)}^2} + {2^2}} \right) - 3{\tan ^{ - 1}}\left( {\frac{{x + 3}}{2}} \right) + C \cr
& = \ln \left( {{x^2} + 6x + 13} \right) - 3{\tan ^{ - 1}}\left( {\frac{{x + 3}}{2}} \right) + C \cr} $$