Answer
$r = -\frac{1}{3}e^{-3t} - e^{-2t}-e^{-t}+C$
Work Step by Step
$\frac{dr}{dt} = \frac{(1+e^{t})^{2}}{e^{3t}}$
Let $e^{t} = x$
$\frac{dr}{dt} = \frac{(1+x)^{2}}{x^{3}}$
$\frac{dr}{dt} = \frac{1+2x+x^{2}}{x^{3}}$
$\frac{dr}{dt} = \frac{1}{x^{3}} + \frac{2x}{x^{3}} + \frac{x^{2}}{x^{3}}$
$\frac{dr}{dt} = x^{-3} + 2x^{-2} + x^{-1}$
$\frac{dr}{dt} = e^{-3t} + 2e^{-2t} + e^{-t}$
$r = \frac{1}{-3}e^{-3t} + (2)\frac{1}{-2}e^{-2t}-e^{-t}+C$
$r = -\frac{1}{3}e^{-3t} - e^{-2t}-e^{-t}+C$
