University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.7 - Improper Integrals - Exercises - Page 471: 9


$\ln 3$

Work Step by Step

Consider $f(x)= \int_{-\infty}^{-2} \dfrac{2}{x^2-1} dx$ Since, we have $\lim\limits_{a \to -\infty} f(x)= \lim\limits_{a \to -\infty}\int_{a}^{-2} [2(\dfrac{1}{x^2-1} dx)]=\lim\limits_{a \to -\infty}\int_{a}^{-2} \dfrac{-1}{x+1}+\dfrac{1}{x-1} $ or, $=\lim\limits_{a \to -\infty} [\ln |x-1|-\ln |x+1|]_{a}^{-2}$ or, $=\lim\limits_{a \to -\infty} \ln 3 -\ln |\dfrac{a-1}{a+1}|$ or, $=\ln 3$
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