University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 61

Answer

$\ln|x|-5\tan^{-1}x+C$

Work Step by Step

$\int (\frac{1}{x}-\frac{5}{x^{2}+1})dx=$$\int \frac{1}{x}dx-5\int\frac{dx}{x^{2}+1^2}=$ $\ln|x|-5\times\frac{1}{1}\tan^{-1}\frac{x}{1}+C$ (As $\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a})$ $=\ln|x|-5\tan^{-1}x+C$
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