Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 476: 10


$$\int\ln\sqrt xdx=x\ln\sqrt x-\frac{x}{2}+C$$

Work Step by Step

$$A=\int\ln\sqrt xdx$$ We would choose $u=\ln\sqrt x$ and $dv=dx$ For $u=\ln\sqrt x$, according to Chain Rule, we have $$du=(\ln\sqrt x)'dx=\frac{d(\ln\sqrt x)}{d(\sqrt x)}\frac{d(\sqrt x)}{dx}dx=(\frac{1}{\sqrt x}\times\frac{1}{2\sqrt x})dx=\frac{1}{2x}dx$$ For $dv=dx$, then $v=x$ Apply Integration by Parts to A, we have $$A=uv−\int vdu$$ $$A=x\ln\sqrt x-\int x\frac{1}{2x}dx$$ $$A=x\ln\sqrt x-\int\frac{1}{2}dx$$ $$A=x\ln\sqrt x-\frac{x}{2}+C$$
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