Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 39


$$\frac{1}{4}\ln \frac{5}{3}. $$

Work Step by Step

Since $u=4t+12$, then $du=4dt$; moreover, when $t:0\to 2$, then $u:12\to 20$ and hence, we have $$ \int_{0}^{2} \frac{d t}{4 t+12}= \frac{1}{4}\int_{12}^{20} \frac{d u}{u}=\frac{1}{4}\ln u|_{12}^{20}\\=\frac{1}{4}(\ln 20-\ln 12)=\frac{1}{4}\ln \frac{20}{12}=\frac{1}{4}\ln \frac{5}{3}. $$
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