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: 11


$\ln 4$

Work Step by Step

We rewrite the integral as: $2 \int_2^{\infty} \dfrac{1}{v^2-v}=\lim\limits_{a \to \infty} \int_2^{a}\dfrac{1}{v-1}-\dfrac{1}{v} dv$ or, $=\lim\limits_{a \to \infty} 2( \ln |a-1| -\ln a)- 2( \ln 1-\ln 2)$ or, $=2 ln 2$ and $ 2\ln 2=\ln 2^2 =\ln 4$
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