Answer
$$\frac{1}{{12}}\ln \left| {\frac{{3 + {x^2}}}{{3 - {x^2}}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{9 - {x^4}}}} dx{\text{ can be written as }}\int {\frac{x}{{9 - {{\left( {{x^2}} \right)}^2}}}} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ }}dx = \frac{1}{{2x}}du \cr
& {\text{Substitute}} \cr
& \int {\frac{x}{{9 - {{\left( {{x^2}} \right)}^2}}}} dx = \int {\frac{x}{{9 - {u^2}}}} \left( {\frac{1}{{2x}}} \right)du \cr
& = \frac{1}{2}\int {\frac{1}{{9 - {u^2}}}} du \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{{a^2} - {u^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{a + u}}{{a - u}}} \right| + C} \cr
& = \frac{1}{2}\left( {\frac{1}{{2\left( 3 \right)}}\ln \left| {\frac{{3 + u}}{{3 - u}}} \right|} \right) + C \cr
& = \frac{1}{{12}}\ln \left| {\frac{{3 + u}}{{3 - u}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{{12}}\ln \left| {\frac{{3 + {x^2}}}{{3 - {x^2}}}} \right| + C \cr} $$
