# 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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