Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 36

Answer

$$\int\frac{2^t}{2^t+3}dt=\frac{\ln(2^t+3)}{\ln2}+C$$

Work Step by Step

$$A=\int\frac{2^t}{2^t+3}dt$$ Let $u=2^t+3$ Then we have $du=2^t\ln2dt$. That means $2^tdt=\frac{1}{\ln2}du$ Substitute into $A$: $$A=\frac{1}{\ln2}\int\frac{1}{u}du$$ $$A=\frac{\ln|u|}{\ln2}+C$$ $$A=\frac{\ln|2^t+3|}{\ln 2}+C$$ As all $t\in R$, $2^t\gt0$. That means $(2^t+3)\gt0$. Therefore, $|2^t+3|=2^t+3$ $$A=\frac{\ln(2^t+3)}{\ln2}+C$$
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