## Calculus 10th Edition

Published by Brooks Cole

# Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises: 10

$\frac{1}{3}$$\ln|x^{3}-3x^{2}| + C #### Work Step by Step \int\frac{x^{2}-2x}{x^{3}-3x^{2}} dx Let u =x^{3}-3x^{2} \frac{du}{dx} = 3x^{2}-6x \frac{du}{3x^{2}-6x} = dx Substitute u and dx into the original equation \int\frac{x^{2}-2x}{u}\frac{du}{3x^{2}-6x} = \int\frac{x^{2}-2x}{3x^{2}-6x}\frac{1}{u} du = \int\frac{x^{2}-2x}{3(x^{2}-2x)}\frac{1}{u} du = \int\frac{1}{3}\frac{1}{u} du = \frac{1}{3}$$\int\frac{1}{u} du$ = $\frac{1}{3}$$\ln|u| + C Since u =x^{3}-3x^{2}, substituting it back will give you \frac{1}{3}$$\ln|x^{3}-3x^{2}|$ + C

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