University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 329: 12

Answer

$$\int12(y^4+4y^2+1)^2(y^3+2y)dy=(y^4+4y^2+1)^3+C$$

Work Step by Step

$$A=\int12(y^4+4y^2+1)^2(y^3+2y)dy$$ We set $u=y^4+4y^2+1$. Then $$du=(4y^3+8y)dy=4(y^3+2y)dy$$ That means $$(y^3+2y)dy=\frac{1}{4}du$$ Therefore, $$A=\int\Big(12u^2\times\frac{1}{4}du\Big)=\int3u^2du$$ $$A=\frac{3u^3}{3}+C=u^3+C$$ $$A=(y^4+4y^2+1)^3+C$$
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