Answer
$$\ln \left| {{y^2} - 25} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{2ydy}}{{{y^2} - 25}}} \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = {y^2} - 25,{\text{ so that }}du = 2ydy \cr
& {\text{write the integral in terms of }}u \cr
& \int {\frac{{2ydy}}{{{y^2} - 25}}} = \int {\frac{{du}}{u}} \cr
& {\text{integrate }} \cr
& = \ln \left| u \right| + C \cr
& {\text{write in terms of }}y{\text{; replace }}{y^2} - 25{\text{ for }}u \cr
& = \ln \left| {{y^2} - 25} \right| + C \cr} $$
