Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Review - True-False Quiz: 5



Work Step by Step

Given: $ \int^{4}_{0}\frac{x}{x^{2}-1}dx$ Since, the $\frac{x}{x^{2}-1}dx$ is bounded at $x=1$ Therefore, $ \int^{4}_{0}\frac{x}{x^{2}-1}dx=\int^{1}_{0}\frac{x}{x^{2}-1}d+ \int^{4}_{1}\frac{x}{x^{2}-1}dx$ $=\frac{1}{2}\int^{1}_{0}\frac{2x}{x^{2}-1}dx+ \frac{1}{2}\int^{4}_{1}\frac{2x}{x^{2}-1}dx$ $=[\frac{1}{2}ln(x^{2}-1]^{1}_{0}+[\frac{1}{2}ln(x^{2}-1]^{4}_{1}$ $=-\infty+\infty$ = Indeterminate Hence, the given integral diverges and is not equal to $\frac{1}{2}ln15$.
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