Answer
$$\frac{1}{2}\ln \left( {{x^2} + 6x + 13} \right) - \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{{x + 3}}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{x^2} + 6x + 13}}} dx \cr
& {\text{completing the square for the denominator}} \cr
& = {x^2} + 6x + {\left( {\frac{6}{2}} \right)^2} - {\left( {\frac{6}{2}} \right)^2} + 13 \cr
& = {x^2} + 6x + 9 + 4 \cr
& = {\left( {x + 3} \right)^2} + {2^2} \cr
& \cr
& \int {\frac{x}{{{x^2} + 6x + 13}}} dx = \int {\frac{x}{{{{\left( {x + 3} \right)}^2} + {2^2}}}} dx \cr
& {\text{make }}u = x + 3,\,\,\,\,{\text{then}}\,\,\,\,\,\,du = dx \cr
& \int {\frac{x}{{{{\left( {x + 3} \right)}^2} + {2^2}}}} dx = \int {\frac{{u - 3}}{{{u^2} + {2^2}}}} du \cr
& = \int {\frac{u}{{{u^2} + {2^2}}}} du - \int {\frac{3}{{{u^2} + {2^2}}}} du \cr
& = \frac{1}{2}\int {\frac{{2u}}{{{u^2} + {2^2}}}} du - 3\int {\frac{1}{{{u^2} + {2^2}}}} du \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral }}\int {\frac{1}{{{u^2} + {2^2}}}} du \cr
& {\text{By formula 68}} \cr
& \left( {68} \right):\,\,\,\,\,\,\int {\frac{{du}}{{{a^2} + {u^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& = \frac{1}{2}\ln \left( {{u^2} + {2^2}} \right) - 3\left( {\frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{u}{2}} \right)} \right) + C \cr
& \cr
& {\text{write in terms of }}x{\text{; replace }}x + 3 {\text{ for }}u \cr
& = \frac{1}{2}\ln \left( {{{\left( {x + 3} \right)}^2} + {2^2}} \right) - 3\left( {\frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{{x + 3}}{2}} \right)} \right) + C \cr
& {\text{then}} \cr
& = \frac{1}{2}\ln \left( {{x^2} + 6x + 13} \right) - 3\left( {\frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{{x + 3}}{2}} \right)} \right) + C \cr
& = \frac{1}{2}\ln \left( {{x^2} + 6x + 13} \right) - \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{{x + 3}}{2}} \right) + C \cr} $$
