University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 43


$$\int\sqrt{x}\ln xdx=\frac{2}{3}x^{3/2}\ln x-\frac{4}{9}x^{3/2}+C$$

Work Step by Step

$$A=\int\sqrt{x}\ln xdx$$ Take $u=\ln x$ and $dv=\sqrt xdx=x^{1/2}dx$ Then we have, $du=\frac{dx}{x}$ and $v=\frac{x^{3/2}}{\frac{3}{2}}=\frac{2x^{3/2}}{3}$ Applying the formula $\int udv=uv-\int vdu$, we have $$A=\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\int x^{3/2}\times\frac{dx}{x}$$ $$A=\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\int x^{1/2}dx$$ $$A=\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\times\frac{2x^{3/2}}{3}+C$$ $$A=\frac{2}{3}x^{3/2}\ln x-\frac{4}{9}x^{3/2}+C$$
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