Answer
$$\int_{0}^{1}\frac{1+12t}{1+3t}dt=4-ln4$$
Work Step by Step
$$\int_{0}^{1}\frac{1+12t}{1+3t}dt=\int_{0}^{1}(4-\frac{3}{1+3t})dt$$
$$=\left[4t-ln(1+3t)\right]_{0}^{1}$$
$$=(4-ln4)-(0-ln1)$$
$$=4-ln4$$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 508: 25
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