Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 328: 79


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

Work Step by Step

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