Answer
$$\frac{1}{2}\ln \left( {{{\left( {x + 1} \right)}^2} + 1} \right) - {\tan ^{ - 1}}\left( {x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{x^2} + 2x + 2}}} dx \cr
& {\text{Completing the square for }}{x^2} + 2x + 2 \cr
& {x^2} + 2x + 2 = \left( {{x^2} + 2x + 1} \right) + 1 \cr
& {x^2} + 2x + 2 = {\left( {x + 1} \right)^2} + 1 \cr
& \int {\frac{x}{{{x^2} + 2x + 2}}} dx = \int {\frac{x}{{{{\left( {x + 1} \right)}^2} + 1}}} dx \cr
& \cr
& {\text{Write in terms of }}u \cr
& {\text{Substitute }}u = x + 1,{\text{ }}\,\,\,du = dx{\text{ and }}x = u - 1 \cr
& \int {\frac{x}{{{{\left( {x + 1} \right)}^2} + 1}}} dx = \int {\frac{{u - 1}}{{{u^2} + 1}}} du \cr
& {\text{Distribute}} \cr
& = \int {\frac{u}{{{u^2} + 1}}} du - \int {\frac{1}{{{u^2} + 1}}} du \cr
& {\text{Integrate}} \cr
& = \frac{1}{2}\ln \left( {{u^2} + 1} \right) - {\tan ^{ - 1}}u + C \cr
& \cr
& {\text{Write in terms of }}x,\,\,{\text{replace }}u = x + 1 \cr
& = \frac{1}{2}\ln \left( {{{\left( {x + 1} \right)}^2} + 1} \right) - {\tan ^{ - 1}}\left( {x + 1} \right) + C \cr} $$
