Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 54

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 $e^{t} = x$ $\frac{dr}{dt} = \frac{(1+x)^{2}}{x^{3}}$ $\frac{dr}{dt} = \frac{1+2x+x^{2}}{x^{3}}$ $\frac{dr}{dt} = \frac{1}{x^{3}} + \frac{2x}{x^{3}} + \frac{x^{2}}{x^{3}}$ $\frac{dr}{dt} = x^{-3} + 2x^{-2} + x^{-1}$ $\frac{dr}{dt} = e^{-3t} + 2e^{-2t} + e^{-t}$ $r = \frac{1}{-3}e^{-3t} + (2)\frac{1}{-2}e^{-2t}-e^{-t}+C$ $r = -\frac{1}{3}e^{-3t} - e^{-2t}-e^{-t}+C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.