## University Calculus: Early Transcendentals (3rd Edition)

$-\dfrac{1}{r+5}+c$
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ where $c$ is a constant of proportionality. Substitute $k=r+5 \implies dr =dk$ Now, $\int \dfrac{dr}{(r+5)^2}dt=\int \dfrac{1}{k^2}dk=\int k^{-2} dk$ or, $=\dfrac{k^{-2+1}}{-2+1}+c$ or, $=\dfrac{k^{-1}}{-1}+c$ plug back in: $u=r+5$ Thus, $\dfrac{k^{-1}}{-1}+c=-\dfrac{1}{r+5}+c$