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 477: 35

Answer

$\displaystyle \frac{32}{5}(\ln 2)^{2}-\frac{64}{25}\ln 2+\frac{62}{125}$

Work Step by Step

Integration by parts: $\displaystyle \int udv=uv-\int vdu$ The idea is to choose a relatively easy $dv$ to integrate, and a u whose $u'$ does not complicate matters (best case: makes thing simpler). ---- $\left[\begin{array}{ll} u=(\ln x)^{2} & dv=x^{4}dx\\ & \\ du=2\frac{\ln x}{x}dx, & v=\frac{1}{5}x^{5} \end{array}\right]$ $I=\displaystyle \int_{1}^{2}x^{4}(\ln x)^{2}dx=uv|_{1}^{2}-\int_{1}^{2}vdu$ $=[\displaystyle \frac{x^{5}}{5}(\ln x)^{2}]_{1}^{2}-\int_{1}^{2}\frac{1}{5}x^{5}\cdot 2\frac{\ln x}{x}dxdx$ $=(\displaystyle \frac{32}{5}(\ln 2)^{2}-0)-\frac{2}{5}\int_{1}^{2}x^{4}\ln xdx$. $=\displaystyle \frac{32}{5}(\ln 2)^{2}-\frac{2}{5}\int_{1}^{2}x^{4}\ln xdx$. By parts, again, $\left[\begin{array}{ll} u=\ln x & dv=x^{4}dx\\ & \\ du=\frac{1}{x}dx & v=\frac{1}{5}x^{5} \end{array}\right]$ $\displaystyle \int_{1}^{2}x^{4}\ln xdx=[\frac{x^{5}}{5}\ln x]_{1}^{2}-\int_{1}^{2}\frac{1}{5}x^{5}\cdot\frac{1}{x}dx==[\frac{x^{5}}{5}\ln x]_{1}^{2}-\frac{1}{5}\int_{1}^{2}x^{4}dx$ $=\displaystyle \frac{32}{5}\ln 2-0-\frac{1}{5}[\frac{x^{5}}{5}]_{1}^{2}$ $=\displaystyle \frac{32}{5}\ln 2-(\frac{32}{25}-\frac{1}{25})$ $=\displaystyle \frac{32}{5}\ln 2-\frac{31}{25}$ $I=\displaystyle \frac{32}{5}(\ln 2)^{2}-\frac{2}{5}(\frac{32}{5}\ln 2-\frac{31}{25})$ $=\displaystyle \frac{32}{5}(\ln 2)^{2}-\frac{64}{25}\ln 2+\frac{62}{125}$.
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