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: 17


$$ \int x^2 \sqrt{x+1} dx =\frac{2}{7}(x+1)^{7/2} -\frac{4}{5}(x+1)^{5/2} +\frac{2}{3}(x+1)^{3/2} +c $$

Work Step by Step

Since $ u=x+1$, then $ du=dx $ and hence, $$ \int x^2 \sqrt{x+1} dx=\int (u-1)^2u^{1/2} d u= \int (u^2-2u+1)u^{1/2} d u\\ =\int u^{5/2} -2u^{3/2} +u^{1/2} d u=\frac{2}{7}u^{7/2} -2\frac{2}{5}u^{5/2} +\frac{2}{3}u^{3/2} +c\\ =\frac{2}{7}(x+1)^{7/2} -\frac{4}{5}(x+1)^{5/2} +\frac{2}{3} (x+1)^{3/2} +c $$
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