Answer
$\frac{1}{2}\ln \left( {{x^2} + 1} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{x^2} + 1}}} dx \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = {x^2} + 1,{\text{ then }}du = 2xdx,{\text{ }}dx = \frac{1}{{2x}}du \cr
& {\text{Applying the substitution}} \cr
& \int {\frac{x}{{{x^2} + 1}}} dx = \int {\frac{x}{u}\left( {\frac{1}{{2x}}} \right)} du \cr
& = \int {\frac{1}{{2u}}} du \cr
& = \frac{1}{2}\int {\frac{1}{u}} du \cr
& {\text{Integrate }} \cr
& = \frac{1}{2}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}{x^2} + 1{\text{ for }}u \cr
& = \frac{1}{2}\ln \left| {{x^2} + 1} \right| + C \cr
& {\text{or}} \cr
& = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C \cr} $$