University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 102


$$\int^{3}_1\frac{(\ln(v+1))^2}{v+1}dv=\frac{(\ln4)^3-(\ln2)^3}{3}=\frac{7 (\ln{2})^3}{3}$$

Work Step by Step

$$A=\int^{3}_1\frac{(\ln(v+1))^2}{v+1}dv$$ We set $u=\ln(v+1)$, which means $$du=\frac{(v+1)'}{v+1}dv=\frac{1}{v+1}dv$$ For $v=1$, we have $u=\ln2$ For $v=3$, we have $u=\ln4$ Therefore, $$A=\int^{\ln4}_{\ln2}u^2du$$ $$A=\frac{1}{3}u^3\Big]^{\ln4}_{\ln2}$$ $$A=\frac{(\ln4)^3-(\ln2)^3}{3}=\frac{7 (\ln{2})^3}{3}$$
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