Answer
$$\frac{1}{8}\ln \left( {25 + 4{x^2}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{xdx}}{{25 + 4{x^2}}}} \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = 25 + 4{x^2},\,\,\,\,du = 8xdx,\,\,\,\,dx = \frac{{du}}{{8x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{xdx}}{{25 + 4{x^2}}}} = \int {\frac{x}{u}\left( {\frac{{du}}{{8x}}} \right)} \cr
& {\text{Cancel the common factor }}x \cr
& = \int {\frac{1}{u}} \left( {\frac{{du}}{8}} \right) \cr
& = \frac{1}{8}\int {\frac{1}{u}du} \cr
& {\text{Integrating, we get:}} \cr
& = \frac{1}{8}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}x.{\text{ Use }}u = 25 + 4{x^2} \cr
& = \frac{1}{8}\ln \left( {25 + 4{x^2}} \right) + C \cr} $$
