Answer
$$\frac{1}{2}\ln \left( {{x^2} + 4x + 8} \right) - {\tan ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{x^2} + 4x + 8}}dx} \cr
& {\text{Completing the square}} \cr
& \int {\frac{x}{{\left( {{x^2} + 4x + 4} \right) + 4}}dx} \cr
& \int {\frac{x}{{{{\left( {x + 2} \right)}^2} + 4}}dx} \cr
& {\text{Let }}u = x + 2,{\text{ }}x = u - 2,{\text{ }}du = dx \cr
& {\text{Substituting}} \cr
& \int {\frac{x}{{{{\left( {x + 2} \right)}^2} + 4}}dx} = \int {\frac{{u - 2}}{{{u^2} + 4}}du} \cr
& = \int {\frac{u}{{{u^2} + 4}}du} - \int {\frac{2}{{{u^2} + 4}}du} \cr
& = \frac{1}{2}\int {\frac{{2u}}{{{u^2} + 4}}du} - \int {\frac{2}{{{u^2} + 4}}du} \cr
& {\text{Integrating by tables}} \cr
& = \frac{1}{2}\ln \left( {{u^2} + 4} \right) - 2\left( {\frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{u}{2}} \right)} \right) + C \cr
& = \frac{1}{2}\ln \left( {{u^2} + 4} \right) - {\tan ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr
& {\text{Write in terms of }}u,{\text{ let }}u = x + 2 \cr
& = \frac{1}{2}\ln \left( {{x^2} + 4x + 8} \right) - {\tan ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right) + C \cr} $$
