Calculus 10th Edition

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

Chapter 6 - Differential Equations - 6.1 Exercises - Page 404: 43

Answer

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

Work Step by Step

$\frac{dy}{dx}$$=$$\frac{x}{1+x^{2}}$ $y=$$\int$$\frac{x}{1+x^{2}}$$dx$ if $u=1+$$x^{2}$, $\frac{1}{2x}$$du$$=$$dx$ $y=$$\int$$\frac{x}{u}$$\times$$\frac{1}{2x}$$du$ $y=$$\int$$\frac{1}{2u}$$du$ $y=$$\frac{1}{2}$$\ln(u)$$+C$ $y=$$\frac{1}{2}$$\ln(1+x^{2})$$+C$
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