Answer
$$\frac{3}{5}{\tan ^{ - 1}}5y + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{3}{{1 + 25{y^2}}}dy} \cr
& {\text{substitute }}u = 5y,{\text{ }}du = 5dy{\text{ and }}dy = \frac{{du}}{5} \cr
& \int {\frac{3}{{1 + 25{y^2}}}dy} = \int {\frac{3}{{1 + {u^2}}}\left( {\frac{{du}}{5}} \right)} \cr
& {\text{take out the constant}} \cr
& = \frac{3}{5}\int {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{find the antiderivative}} \cr
& = \frac{3}{5}{\tan ^{ - 1}}u + C \cr
& {\text{ with}}\,\,\,u = 5y \cr
& = \frac{3}{5}{\tan ^{ - 1}}5y + C \cr} $$
