Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 346: 6


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

Work Step by Step

To solve the integral $\int \sqrt{2t+1}dt$ we will use substitution $u=2t+1$ which gives us $du=2dt\Rightarrow\frac{1}{2}du=dt.$ Putting this into the integral we get: $$\int \sqrt{2t+1}dt=\int\sqrt{u}\cdot\frac{1}{2}du=\frac{1}{2}\int u^{1/2}du=\frac{1}{2}\frac{u^{3/2}}{\frac{3}{2}}+c=\frac{1}{2}\frac{2}{3}u^{3/2}+c=\frac{u^{3/2}}{3}+c$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $t$ by expressing $u$ in terms of $t$: $$\int \sqrt{2t+1}dt=\frac{u^{3/2}}{3}+c=\frac{(2t+1)^{3/2}}{3}+c$$
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