Calculus: Early Transcendentals 8th Edition

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

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises: 41

Answer

$\ln\frac{\sqrt x}{x+1}$

Work Step by Step

$\frac{1}{3}\ln(x+2)^3+\frac{1}{2}[\ln x-\ln(x^2+3x+2)^2]=\ln(x+2)^{3\times\frac{1}{3}}+\ln x^\frac{1}{2}-\ln(x^2+3x+2)^{2\times\frac{1}{2}}=\ln(x+2)+\ln x^\frac{1}{2}-\ln(x^2+3x+2)=\ln\frac{x^\frac{1}{2}(x+2)}{x^2+3x+2}=\ln\frac{x^\frac{1}{2}(x+2)}{(x+1)(x+2)}=\ln\frac{\sqrt x}{x+1}$
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