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 395: 22


$$\int x^{-5}lnx\,dx=-\frac{lnx}{4x^{4}}-\frac{1}{16x^{4}}+C$$

Work Step by Step

$${(-\frac{x^{-4}lnx}{4})}'=x^{-5}lnx-\frac{x^{-5}}{4}$$ $$\int x^{-5}lnx\,dx=-\frac{x^{-4}lnx}{4}+\frac{1}{4}\int x^{-5}dx$$ $$=-\frac{lnx}{4x^{4}}-\frac{1}{16x^{4}}+C$$
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