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: 109



Work Step by Step

Let $u=t-3$, then $du= dt$ and hence we get $$\int t^2\sqrt{t-3}dt= \int (u+3)^2u^{1/2}du \\ =\int (u^{5/2}+6u^{3/2}+9u^{1/2})du\\ =\frac{1}{7/2}u^{7/2}+\frac{6}{5/2}u^{5/2}+\frac{9}{3/2}u^{3/2}+c\\ =\frac{2}{7}u^{7/2}+\frac{12}{5}u^{5/2}+\frac{18}{3}u^{3/2}+c\\ =\frac{2}{7}(t-3)^{7/2}+\frac{12}{5}(t-3)^{5/2}+6(t-3)^{3/2}+c.$$
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