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: 44


$$ \frac{1}{42}\left(z^{3}+1\right)^{14}+\frac{1}{13}\left(z^{3}+1\right)^{13}+c$$

Work Step by Step

Given $$ \int\left(z^{5}+4 z^{2}\right)\left(z^{3}+1\right)^{12} d z= \int z^{2}\left(z^{3}+4\right)\left(z^{3}+1\right)^{12} d z$$ Let $$u=z^{3}+1 \ \ \ \Rightarrow \ \ du= (3z^2 )dz $$ then \begin{align*} \int\left(z^{5}+4 z^{2}\right)\left(z^{3}+1\right)^{12} d z &=\frac{1}{3} \int(u+3) u^{12} d u \\ &=\frac{1}{3} \int\left(u^{13}+3 u^{12}\right) d u \\ &=\frac{1}{3}\left(\frac{1}{14} u^{14}+3\left(\frac{1}{13} u^{13}\right)+c\right) \\ &=\frac{1}{42} u^{14}+\frac{1}{13} u^{13}+c\\ &=\frac{1}{42} u^{14}+\frac{1}{13} u^{13}+c\\ &= \frac{1}{42}\left(z^{3}+1\right)^{14}+\frac{1}{13}\left(z^{3}+1\right)^{13}+c \end{align*}
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