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: 57



Work Step by Step

$\displaystyle{x^2\ln x=4\ln x}\\ \displaystyle{x^2\ln x-4\ln x=0}\\ \displaystyle{\ln x\left(x^2-4\right)=0}\\ \displaystyle{\ln x=0 \qquad x^2-4=0}\\ \displaystyle{e^{\ln x}=e^0 \qquad x^2=4}\\ \displaystyle{x=e^0 \qquad x=\pm2\qquad x\gt0}\\ \displaystyle{x=1 \qquad x=2}\\ $ $\displaystyle{I=\int_{1}^{2}4\ln x-x^2\ln x\ dx}\\ \displaystyle{I=\int_{1}^{2}\ln x\left(x^2-4\right)\ dx}\\ $ $\displaystyle \left[\begin{array}{ll} u=\ln x & dv=4-x^2 \\ & \\ du=\frac{1}{x} & v=4x-\frac{1}{3}x^3 \end{array}\right]$ Integration by parts $\displaystyle{I=\left[\ln x\times4x-\frac{1}{3}x^3\right]_1^2-\int_{1}^{2}\left(4x-\frac{1}{3}x^3\right)\left(\frac{1}{x}\right)\ dx}\\ \displaystyle{I=\frac{16}{3}\ln2-\int_{1}^{2}4-\frac{1}{3}x^2\ dx}\\ \displaystyle{I=\frac{16}{3}\ln2-\left[4x-\frac{1}{9}x^3\right]_{1}^{2}}\\ \displaystyle{I=\frac{16}{3}\ln2-\frac{29}{9}}\\$
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