Answer
$$\int e^{-5r}dr=-\frac{1}{5}e^{-5r}+c$$
Work Step by Step
To evaluate the integral $$\int e^{-5r}dr$$
we will use substitution $-5r=t$ which gives us $-5dr=dt\Rightarrow dr=-\frac{dt}{5}$. Putting this into the integral we get:
$$\int e^{-5r}dr=\int e^t \Big(-\frac{dt}{5}\Big)=-\frac{1}{5}\int e^tdt=-\frac{1}{5}e^t+c$$
where $c$ is arbitrary constant. Now we have to express solution in terms of $x$:
$$\int e^{-5r}dr=-\frac{1}{5}e^t+c=-\frac{1}{5}e^{-5r}+c$$