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: 5


$$\int \frac{x^3}{x^4-5} dx=\frac{\ln|x^4-5|}{4}+C$$

Work Step by Step

$$A=\int \frac{x^3}{x^4-5} dx$$ Let $u=x^4-5$ Then $du=(x^4-5)'dx=4x^3dx$. So $x^3dx=\frac{1}{4}du$ Substitute into $A$, we have $$A=\int \frac{1}{u}(\frac{1}{4})du$$ $$A=\frac{1}{4}\int\frac{1}{u}du$$ $$A=\frac{1}{4}\ln|u|+C$$ $$A=\frac{\ln|x^4-5|}{4}+C$$
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