Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.4 Integration of Rational Functions by Partial Fractions - 7.4 Exercises - Page 541: 51

Answer

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

Work Step by Step

Given $$\int \frac{d x}{1+e^{x}}$$ \begin{array}{l}{\text { Let } u=e^{x}, \ \ d u=e^{x} d x \ \text { and } d x=\frac{d u}{u} \text { . Then } \\ \begin{aligned}I&=\int \frac{d x}{1+e^{x}}\\ &=\int \frac{d u}{(1+u) u}\\ \text{Since} \ \ \ \frac{1}{u(u+1)} &=\frac{A}{u}+\frac{B}{u+1}\end{aligned} \\ \Rightarrow} \\ {1=A(u+1)+B u \\ \text { Setting } u=-1 \Rightarrow B=-1 . \text { Setting } u=0\Rightarrow A=1. \text { Thus, }} \\ {\begin{aligned}I&=\int \frac{d u}{u(u+1)}\\&=\int\left(\frac{1}{u}-\frac{1}{u+1}\right) d u\\&=\ln |u|-\ln |u+1|+C\\ &=\ln e^{x}-\ln \left(e^{x}+1\right)+C\\ &=x-\ln \left(e^{x}+1\right)+C \end{aligned}}\end{array}
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