Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 534: 22

Answer

$1$

Work Step by Step

Table of integrals : 101. $\displaystyle \int u^{n}\ln udu=\frac{u^{n+1}}{(n+1)^{2}}[(n+1)\ln u-1]+C$ $\displaystyle \int x^{-2}\ln xdx=\frac{x^{-1}}{(-1)^{2}}[(-1)\ln x-1]+C =\displaystyle \frac{-\ln x-1}{x}+C$ This is a type I improper integral. $\displaystyle \int_{1}^{\infty}\frac{\ln x}{x^{2}}dx=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{\ln x}{x^{2}}dx=\lim_{t\rightarrow\infty}\left[\frac{-\ln x-1}{x}\right]_{1}^{t}$ $=\displaystyle \lim_{t\rightarrow\infty}\frac{-\ln t-1}{t}-(-1)$ ...Apply L'Hospital's rule, $\displaystyle \frac{\infty}{\infty}$ $=\displaystyle \lim_{t\rightarrow\infty}\frac{-\frac{1}{t}}{1}+1$ $=0+1$ $=1$
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