Calculus 10th Edition

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

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 512: 40

Answer

$\frac 1 {10} arctan(\frac {2x} 5)+C$

Work Step by Step

$\int \frac 1 {25+4x^2}dx$ $a^2+25, u^2=4x^2$ $a=5, u=2x, \frac 1 2 du=dx$ $\frac 1 2 \int \frac {du}{a^2+u^2}$ $\frac 1 2 \times \frac 1 5 arctan(\frac {2x}5)+C$ $\frac 1 {10} arctan(\frac {2x} 5)+C$
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