University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 37

Answer

$$\int\frac{\log_{10}x}{x}dx=\frac{(\ln x)^2}{2\ln10}+C$$

Work Step by Step

$$A=\int\frac{\log_{10}x}{x}dx=\int\frac{\frac{\ln x}{\ln10}}{x}dx$$ $$A=\frac{1}{\ln10}\int\frac{\ln x}{x}dx$$ We set $u=\ln x$, which means $$du=\frac{1}{x}dx$$ Therefore, $$A=\frac{1}{\ln10}\int udu$$ $$A=\frac{1}{\ln10}\times\frac{u^2}{2}+C$$ $$A=\frac{u^2}{2\ln10}+C=\frac{(\ln x)^2}{2\ln10}+C$$
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