Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 11


$$ \int t\sqrt{t^2+1} dt =\frac{1}{3} (t^2+1)^{3/2}+c. $$

Work Step by Step

Since $ u=t^2+1 $, then $ du=2tdt $ and hence, $$ \int t\sqrt{t^2+1} dt=\frac{1}{2}\int u^{1/2} d u=\frac{1}{2} \frac{2}{3}u^{3/2}+c\\ =\frac{1}{3} (t^2+1)^{3/2}+c. $$
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