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: 43


$y=2\ln|x-9|+C$ When C=$4-2\ln 9$, the graph passes through $(4,0)$

Work Step by Step

$\displaystyle \frac{2x}{x^{2}-9x}=\frac{2x}{x(x-9)}=\frac{2}{x-9 }, x\neq 0$ $y=\displaystyle \int\frac{2}{x-9 }dx=2\int\frac{1}{x-9 }dx\qquad\left[\begin{array}{l} u=x-9\\ du=dx \end{array}\right]$ $=2\displaystyle \int\frac{1}{u}du$ $=2\ln|u|+C$ $y=2\ln|x-9|+C$ $(x,y)=(0, 4)$ is on the graph, so we find C $4=2\ln|0-9|+C$ $C=4-2\ln 9$
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