Answer
$\frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{{x^2}}}{3}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{x^4} + 9}}} dx \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{x}{{{x^4} + 9}}} dx = \int {\frac{x}{{{{\left( {{x^2}} \right)}^2} + 9}}} dx \cr
& {\text{Let }}t = {x^2},{\text{ }}dt = 2xdx \cr
& {\text{Substituting}} \cr
& \int {\frac{x}{{{{\left( {{x^2}} \right)}^2} + 9}}} dx = \int {\frac{{\left( {1/2} \right)}}{{{t^2} + 9}}} dt \cr
& = \frac{1}{2}\int {\frac{1}{{{t^2} + 9}}} dt \cr
& {\text{Use the formula }}\int {\frac{1}{{{u^2} + {a^2}}}du = } \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& = \frac{1}{2}\left( {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{t}{3}} \right)} \right) + C \cr
& = t\ln t - \int {dt} \cr
& = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{t}{3}} \right) + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}{x^2}{\text{ for }}t \cr
& = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{{x^2}}}{3}} \right) + C \cr} $$
