Answer
$-\dfrac{1}{r+5}+c$
Work Step by Step
Use formula $\int x^{a} dx=\dfrac{x^{(a+1)}}{(a+1)}+c$
where $c$ refers to a constant of proportionality.
Plug $a=r+5$ and $dr =da$
Then, we get $\int \dfrac{dr}{(r+5)^2}dt=\int \dfrac{1}{a^2}da=\int a^{-2} da=\dfrac{a^{-2+1}}{-2+1}+c$
or, $=\dfrac{k^{-1}}{-1}+c$
Thus, we have $\dfrac{a^{-1}}{-1}+c=-\dfrac{1}{r+5}+c$
