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 429: 66


See below for detailed work.

Work Step by Step

$$A=\int(\ln x)^ndx$$ We set $u= (\ln x)^n$ and $dv=dx$ That makes $du=\frac{n(\ln x)^{n-1}}{x}dx$ and $v=x$ Applying integration by parts $\int udv=uv-\int vdu$, we have $$A=x(\ln x)^n-\int x\times\frac{n(\ln x)^{n-1}}{x}dx$$ $$A=x(\ln x)^n-n\int(\ln x)^{n-1}dx$$ The reduction formula has been established.
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