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: 2

Answer

$= \frac{2}{3}\sqrt {x^{3}}(\ln x - \frac{2}{3}) + C$

Work Step by Step

$\int \sqrt x \ln x dx$ $u = \ln x$ $u' = \frac{1}{x}$ $\frac{du}{dx} = \frac{1}{x}$ $du = \frac{dx}{x}$ $dv = \sqrt x dx$ $v = \frac{x^{\frac{3}{2}}}{\frac{3}{2}}$ $v = \frac{2}{3}x^{\frac{3}{2}}$ $uv - \int vdu$ $= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \int (\frac{2}{3}x^{\frac{3}{2}})(\frac{dx}{x})$ $= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{2}{3}\int (x^{\frac{3}{2}})(\frac{1}{x})dx$ $= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{2}{3}\int (x^{\frac{3}{2}})(x^{-1})dx$ $= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{2}{3}\int (x^{\frac{1}{2}})dx$ $= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{2}{3}\frac{2}{3}x^{\frac{3}{2}}$ $= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{4}{9}x^{\frac{3}{2}}$ $= \frac{2}{3}x^{\frac{3}{2}}(\ln x - \frac{2}{3}) + C$ $= \frac{2}{3}\sqrt {x^{3}}(\ln x - \frac{2}{3}) + C$
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