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 - Review Exercises - Page 393: 4



Work Step by Step

$\ln[(x^2+1)(x-1)] = \ln\left[\frac{(x^2+1)(x-1)(x+1)}{x+1}\right]= \ln\left[\frac{(x^2+1)(x^2-1)}{x+1}\right] = \ln\left[\frac{x^4-1}{x+1}\right]=\boxed{\ln(x^4-1)-\ln(x+1)}$
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