Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 20

Answer

$$\int\frac{z^2}{z^3+1}dz=\frac{\ln|z^3+1|}{3}+C$$

Work Step by Step

$$A=\int\frac{z^2}{z^3+1}dz$$ Let $u=z^3+1$. Then $du=3z^2dz$, so $z^2dz=\frac{1}{3}du$ Substitute into $A$, we have $$A=\frac{1}{3}\int\frac{1}{u}du$$ $$A=\frac{1}{3}\ln|u|+C$$ $$A=\frac{\ln|z^3+1|}{3}+C$$
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