Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises: 13

Answer

$F(x) = \frac{x}{5} - 2\ln|x| + C$, $x\ne0$

Work Step by Step

$f(x) = \frac{1}{5} - \frac{2}{x}$ Rewrite the second term as a product. $f(x) = \frac{1}{5} - 2(\frac{1}{x})$ Rewriting the second term makes it easy to take the antiderivative (since we already know the antiderivative of $\frac{1}{x}$). $F(x) = \frac{1}{5} - 2\ln|x|$ Now we take the antiderivative of the first term and add the constant. $F(x) = \frac{x}{5} - 2\ln|x| + C$, $x\ne 0$
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