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 - 5.5 Exercises: 77

Answer

$\frac{ln(3^{2x}+1)}{2ln(3)}+C$

Work Step by Step

let u=$3^{2x}+1$ $2ln(3)3^{2x}$ $dx=du$ $dx=\frac{1}{2ln(3)3^{2x}}du$ $\int\frac{3^{2x}}{1+3^{2x}}dx$ $=\frac{1}{2ln(3)}\int \frac{1}{u}du$ $=\frac{ln(u)}{2ln(3)}+C$ $=\frac{ln(3^{2x}+1)}{2ln(3)}+C$
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