Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 396: 83


$$x(\ln x)^{3}-3 x(\ln x)^{2}+6 x \ln x-6 x+C$$ $$x(\ln x)^{2}-2 x(\ln x) +2 x+C$$

Work Step by Step

Since $$\int(\ln x)^{k} d x=x(\ln x)^{k}-k \int(\ln x)^{k-1} d x$$ Then for $k=2$ \begin{align*} \int(\ln x)^{2} d x&=x(\ln x)^{2}-2 \int(\ln x) d x\\ &=x(\ln x)^{2}-2\left(x(\ln x) - \int d x \right)\\ &=x(\ln x)^{2}-2 x(\ln x) +2 x+C \\ \end{align*} For $k=3$ \begin{align*} \int(\ln x)^{3} d x&=x(\ln x)^{3}-3 \int(\ln x)^2 d x\\ &=x(\ln x)^{3}-3\left( x(\ln x)^{2}-2 x(\ln x) +2 x \right)\\ &= x(\ln x)^{3}-3 x(\ln x)^{2}+6 x \ln x-6 x+C\\ \end{align*}
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