Answer
$ - \frac{1}{5}{e^{ - 5r}} + C$
Work Step by Step
$$\eqalign{
& \int {{e^{ - 5r}}} dr \cr
& {\text{Let }}u = - 5r,{\text{ then }}du = - 5dr,{\text{ }}dr = - \frac{1}{5}du \cr
& {\text{Apply the substitution}} \cr
& \int {{e^{ - 5r}}} dr = \int {{e^u}} \left( { - \frac{1}{5}du} \right) \cr
& = - \frac{1}{5}\int {{e^u}} du \cr
& {\text{Integrate}} \cr
& = - \frac{1}{5}{e^u} + C \cr
& {\text{Write in terms of }}r,{\text{ substitute }}u = - 5r \cr
& = - \frac{1}{5}{e^{ - 5r}} + C \cr} $$
