University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 101


$$\int^{e}_1\frac{1}{x}(1+7\ln x)^{-1/3}dx=\frac{9}{14}$$

Work Step by Step

$$A=\int^{e}_1\frac{1}{x}(1+7\ln x)^{-1/3}dx$$ We set $u=1+7\ln x$, which means $$du=\frac{7}{x}dx$$ $$\frac{1}{x}dx=\frac{1}{7}du$$ For $x=e$, we have $$u=1+7\ln e=1+7=8$$ For $x=1$, we have $$u=1+7\ln1=1+7\times0=1$$ Therefore, $$A=\frac{1}{7}\int^8_1u^{-1/3}du$$ $$A=\frac{1}{7}\times\frac{3u^{2/3}}{2}\Big]^8_1=\frac{3}{14}u^{2/3}\Big]^8_1$$ $$A=\frac{3}{14}(8^{2/3}-1)$$ $$A=\frac{3}{14}(4-1)=\frac{9}{14}$$
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