University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Practice Exercises - Page 420: 12


$$\int^{4}_2(1+\ln t)t\ln tdt=30(\ln2)^2$$

Work Step by Step

$$A=\int^{4}_2(1+\ln t)t\ln tdt$$ Set $u=t\ln t$, which means $$du=(t'\ln t+t(\ln t)')dt=(\ln t+\frac{t}{t})dt$$ $$du=(\ln t+1)dt$$ - For $t=2$, we have $u=2\ln2$ - For $t=4$, we have $u=4\ln4$ Therefore, $$A=\int^{4\ln4}_{2\ln2}udu=\frac{u^2}{2}\Big]^{4\ln4}_{2\ln2}$$ $$A=\frac{1}{2}\Big(16(\ln4)^2-4(\ln2)^2\Big)$$ $$A=8(\ln4)^2-2(\ln2)^2$$ $$A=8(\ln2^2)^2-2(\ln2)^2$$ $$A=8(2\ln2)^2-2(\ln2)^2$$ $$A=32(\ln2)^2-2(\ln2)^2=30(\ln2)^2$$
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