Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 501: 47


$$\int \frac{e^{2 x}}{e^{2 x}+3 e^{x}+2} d x =\ln \frac{\left(e^{x}+2\right)^{2}}{e^{x}+1}+C $$

Work Step by Step

Given $$\int \frac{e^{2 x}}{e^{2 x}+3 e^{x}+2} d x$$ \begin{array}{l}{\text { Let } u=e^{x} \text { . Then } x=\ln u, d x=\frac{d u}{u} \Rightarrow} \\ {\qquad \begin{aligned} I&=\int \frac{e^{2 x}}{e^{2 x}+3 e^{x}+2} d x\\ &=\int \frac{u^{2}(d u / u)}{u^{2}+3 u+2}\\ &=\int \frac{u d u}{(u+1)(u+2)}\\ \end{aligned}}\end{array} Since we have \begin{array}{l}{ \frac{u }{(u+1)(u+2)} =\frac{A}{(u+1) }+\frac{B}{ (u+2)} }\\{\text{this implies}}\\{ u=A(u+2)+B(u+1) }\\{\text{at} \ u=-1 \rightarrow A=-1}\\{\text{at} \ u=-2 \rightarrow B=2}\\{ }\end{array} So, we get \begin{aligned} I&=\int\left[\frac{-1}{u+1}+\frac{2}{u+2}\right] d u \\ &=2 \ln |u+2|-\ln |u+1|+C\\ &= \ln \frac{|u+2|^2}{|u+1|}+C\\ &=\ln \frac{\left(e^{x}+2\right)^{2}}{e^{x}+1}+C \end{aligned}
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.