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