University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 36


$$\int \frac{(\ln x)^3}{x}dx=\frac{(\ln x)^4}{4}+C$$

Work Step by Step

$$A=\int \frac{(\ln x)^3}{x}dx$$ We set $u=\ln x$. We then have $$du=\frac{dx}{x}$$ Therefore, $$A=\int u^3du$$ $$A=\frac{u^4}{4}+C$$ $$A=\frac{(\ln x)^4}{4}+C$$
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