Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 78



Work Step by Step

We have $$\int_0^2 \frac{dt}{4t+12}=\frac{1}{4}\int_0^2 \frac{4dt}{4t+12}=\frac{1}{4}\ln (4t+12)|_0^2\\ =\frac{1}{4}(\ln 20 -\ln 12)=\frac{1}{4}\ln \frac{20}{12}=\frac{1}{4}\ln\frac{5}{3}.$$
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