Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 391: 107

Answer

\begin{aligned} \int_{1}^{\ln x} \frac{1}{t} d t =\ln (\ln x), \ \ \ \ \ \ \ \ x>1 \end{aligned}

Work Step by Step

Given $$\int_{1}^{\ln x} \frac{1}{t} d t $$ So, we have \begin{aligned} I&=\int_{1}^{\ln x} \frac{1}{t} d t\\ &=[\ln |t|]_{1}^{\ln x}\\ &=\ln |\ln x|-\ln 1\\ &=\ln (\ln x), \ \ \ \ \ \ \ \ x>1 \end{aligned}
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