University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 25


$\ln (1+e^r)+c$

Work Step by Step

Solve $\int \dfrac{e^r}{1+e^r} dr $ Let us $p=1+e^r$ and $dp=e^{r}dr$ This implies $\int \dfrac{e^r}{1+e^r} dr=\int\dfrac{1}{p}dp$ $=\ln |p|+c$ $=\ln (1+e^r)+c$ Hence, $\int \dfrac{e^r}{1+e^r} dr=\ln (1+e^r)+c$
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