Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 6


$-\frac{9}{4}$$\ln|5-4x|$ + C

Work Step by Step

$\int\frac{9}{5-4x} dx $ Let $u = 5-4x$ $\frac{du}{dx}$ = -4 $\frac{du}{-4}$ = $dx$ Substitute $u$ and $dx$ into the original equation $\int\frac{9}{u}\frac{du}{-4}$ = $\int\frac{9}{-4}\frac{1}{u} du$ = $-\frac{9}{4}$$\int\frac{1}{u} du$ = $-\frac{9}{4}$$\ln|u|$ + C Since $u = 5-4x$, substituting it back will give you $-\frac{9}{4}$$\ln|5-4x|$ + C
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