Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 426: 42

Answer

$\frac{1}{{\ln 2}}\ln \left| {{2^t} + 3} \right| + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{{2^t}}}{{{2^t} + 3}}} dt \cr & {\text{Let }}u = {2^t} + 3,{\text{ }}du = {2^t}\left( {\ln 2} \right)dt,{\text{ }}\frac{1}{{\ln 2}}du = {2^t}dt \cr & {\text{Applying the substitution}} \cr & \int {\frac{{{2^t}}}{{{2^t} + 3}}} dt = \int {\frac{1}{u}} \left( {\frac{1}{{\ln 2}}} \right)du \cr & = \frac{1}{{\ln 2}}\int {\frac{1}{u}} du \cr & {\text{Integrate }} \cr & = \frac{1}{{\ln 2}}\ln \left| u \right| + C \cr & {\text{Write in terms of }}t,{\text{ substitute }}u = {2^t} + 3 \cr & = \frac{1}{{\ln 2}}\ln \left| {{2^t} + 3} \right| + C \cr} $$
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