Answer
$$\ln \left| {4{r^2} - 5} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{8rdr}}{{4{r^2} - 5}}} \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = 4{r^2} - 5,{\text{ so that }}du = 8rdr \cr
& {\text{Write the integral in terms of }}u \cr
& \int {\frac{{8rdr}}{{4{r^2} - 5}}} = \int {\frac{{du}}{u}} \cr
& {\text{integrate }} \cr
& = \ln \left| u \right| + C \cr
& {\text{write in terms of }}y{\text{; replace }}4{r^2} - 5{\text{ for }}u \cr
& = \ln \left| {4{r^2} - 5} \right| + C \cr} $$
