University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 57


$$\int \frac{dz}{1+e^z}=z-\ln(1+e^z)+C$$

Work Step by Step

$$A=\int \frac{dz}{1+e^z}$$ We set $u=e^z$ Then $$du=e^zdz=udz$$ $$dz=\frac{1}{u}du$$ Therefore, $$A=\int\frac{\frac{1}{u}}{1+u}du=\int\frac{1}{u(1+u)}du$$ $$A=\int\frac{(1+u)-(u)}{u(1+u)}du=\int\Big(\frac{1+u}{u(1+u)}-\frac{u}{u(1+u)}\Big)du$$ $$A=\int\Big(\frac{1}{u}-\frac{1}{1+u}\Big)du$$ $$A=\ln|u|-\ln|1+u|+C$$ $$A=\ln|e^z|-\ln|1+e^z|+C$$ Since $e^z\gt0$ for all $z$, we have $|e^z|=e^z$ and $|1+e^z|=1+e^z$ $$A=\ln e^z-\ln (1+e^z)+C$$ $$A=z-\ln(1+e^z)+C$$
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