University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 56


$$\int \frac{\ln\sqrt t}{t}dt=\frac{\ln^2t}{4}+C$$

Work Step by Step

$$A=\int \frac{\ln\sqrt t}{t}dt=\int\frac{\ln t^{1/2}}{t}dt$$ $$A=\frac{1}{2}\int\frac{\ln t}{t}dt$$ We set $u=\ln t$ Then $$du=\frac{1}{t}dt$$ Therefore, $$A=\frac{1}{2}\int udu=\frac{1}{2}\times\frac{u^2}{2}+C$$ $$A=\frac{u^2}{4}+C$$ $$A=\frac{\ln^2t}{4}+C$$
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