Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 50


$\displaystyle \frac{1}{2}\ln 5\approx 0.805$ check with desmos online calculator:

Work Step by Step

$I=\displaystyle \int_{-1}^{1}\frac{1}{2x+3}dx=$ Find the indefinite integral first, $\displaystyle \int\frac{1}{2x+3}dx=\left[\begin{array}{ll} u=2x+3 & \\ du=2dx & dx=\frac{1}{2}du \end{array}\right]$ $=\displaystyle \frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}\ln|u|+C$ $=\displaystyle \frac{1}{2}\ln|2x+3|+C$ Now, the definite integral: $I=\left[\displaystyle \frac{1}{2}\ln|2x+3|\right]_{-1}^{1}$ $=\displaystyle \frac{1}{2}(\ln 5-\ln 1)$ $=\displaystyle \frac{1}{2}\ln 5$ $\approx 0.805$
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