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: 82


\begin{align*} \int(\ln x)^{k} d x&=x(\ln x)^{k}-k \int(\ln x)^{k-1} d x \end{align*}

Work Step by Step

Given $$\int(\ln x)^{k} d x$$ Let \begin{align*} u&= (\ln x)^k \ \ \ \ \ \ \ dv=dx\\ du& =\frac{k}{x}(\ln x)^{k-1}\ \ \ \ \ v=x \end{align*} Then \begin{align*} \int(\ln x)^{k} d x&=x(\ln x)^{k}-k \int(\ln x)^{k-1} d x \end{align*}
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