## Calculus 10th Edition

Published by Brooks Cole

# Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 42

#### Answer

$y=x-2\ln|x|+1$ When C=1, the graph passes through $(-1,0)$

#### Work Step by Step

$\displaystyle \frac{x-2}{x}=\frac{x}{x}-\frac{2}{x}=1-\frac{2}{x}$, $y=\displaystyle \int(1-\frac{2}{x})dx=\int dx-2\int\frac{dx}{x}$ $y=x-2\ln|x|+C$ $(x,y)=(-1, 0)$ is on the graph, so we find C $0=-1-2\ln|-1|+C$ $0=-1+0+C$ $C=1$ $y=x-2\ln|x|+1$

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