Answer
$$ - \frac{1}{{18}}\ln \left| {25 - 9{x^2}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{25 - 9{x^2}}}} dx \cr
& {\text{set }}u = 25 - 9{x^2}{\text{ then }}\frac{{du}}{{dx}} = - 18x \to dx = \frac{{du}}{{ - 18x}} \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{x}{{25 - 9{x^2}}}} dx = \int {\frac{x}{u}} \left( {\frac{{du}}{{ - 18x}}} \right) \cr
& {\text{cancel the common factors}} \cr
& = \int {\frac{1}{u}} \left( {\frac{{du}}{{ - 18}}} \right) \cr
& = - \frac{1}{{18}}\int {\frac{1}{u}} du \cr
& {\text{integrating}} \cr
& = - \frac{1}{{18}}\ln \left| u \right| + C \cr
& {\text{replace }}25 - 9{x^2}{\text{ for }}u \cr
& = - \frac{1}{{18}}\ln \left| {25 - 9{x^2}} \right| + C \cr} $$
