Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 86


$$\frac{2}{3}(e^t+1)^{3/2}+c $$

Work Step by Step

Let $u=e^t+1$, then $du=e^t dt$. Now, we have $$ \int e^{t} \sqrt{e^{t}+1} d t=\int u^{1/2}du\\ =\frac{1}{3/2} u^{3/2}+c=\frac{2}{3}(e^t+1)^{3/2}+c $$
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