Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises - Page 579: 5


$\frac{1}{2}((\ln(2e))^{2})-(\ln(2)^2)$ Or $\approx1.193$

Work Step by Step

$\int_{1}^{e}\frac{\ln(2x)}{x}dx$ Use u-subsitution. $\frac{du}{dx}= \ln(2x)$ $\frac{du}{dx}=\frac{1}{2x}(2)$ $dx=xdu$ $\int_{1}^{e}\frac{ux}{x}du$ $\int_{1}^{e}udu$ $\frac{u^{2}}{2}|_{1}^{e}$ $\frac{(\ln(2x))^{2}}{2}|_{1}^{e}$ $(\frac{(\ln(2e))^{2}}{2})-(\frac{(\ln(2))^{2}}{2})$ $\frac{1}{2}((\ln(2e))^{2})-(\ln(2)^2)$ Or $\approx1.193$
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