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 - Page 418: 6


$$\int\sqrt{2t+1}dt=\frac{1}{3}\sqrt{(2t+1)^3} + C$$

Work Step by Step

$$A=\int\sqrt{2t+1}dt$$ Let $u=2t+1$ Then $du=(2t+1)'dt=2dt$. So $dt=\frac{1}{2}du$ Substitute into $A$, we have $$A=\int \sqrt u(\frac{1}{2})du$$ $$A=\frac{1}{2}\int u^{1/2}du$$ $$A=\frac{1}{2}\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{\sqrt{u^3}}{3}+C$$ $$A=\frac{1}{3}\sqrt{(2t+1)^3} + C$$
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