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: 10

Answer

$\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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