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: 41

Answer

$y=-3\ln|x-2|+C$ when $C=0$, the graph passes through (1,0)
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Work Step by Step

$y=\displaystyle \int\frac{3}{2-x}dx =-3\displaystyle \int\frac{1}{x-2}dx$ $\left[\begin{array}{l} u=x-2\\ du=dx \end{array}\right]$ $=-3\displaystyle \int\frac{1}{u}du$ $=-3\ln|u|+C$ $y=-3\ln|x-2|+C$ $(x,y)=(1, 0)$ is on the graph, so we find C $0=-3\ln|1-2|+C$ $0=-3\ln 1+C$ $C=0$ $y=-3\ln|x-2|$
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