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: 92



Work Step by Step

Let $u=t+1$, then $du=dt$. Now, we have $$\int (t+1)\sqrt{t+1}dt= \int u \sqrt u du=\int u^{3/2}du\\ =\frac{1}{5/2}u^{5/2}+c=\frac{2}{5}(t+1)^{5/2}+c$$
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