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: 7


$\frac{1}{2}$$\ln|x^{2}-3|$ + C

Work Step by Step

$\int\frac{x}{x^{2}-3} dx $ Let $u =x^{2}-3$ $\frac{du}{dx}$ = $2x$ $\frac{du}{2x}$ = $dx$ Substitute $u$ and $dx$ into the original equation $\int\frac{x}{u}\frac{du}{2x}$ = $\int\frac{x}{2x}\frac{1}{u} du$ = $\int\frac{1}{2}\frac{1}{u} du$ = $\frac{1}{2}$$\int\frac{1}{u} du$ = $\frac{1}{2}$$\ln|u|$ + C Since $u =x^{2}-3$, substituting it back will give you $\frac{1}{2}$$\ln|x^{2}-3|$ + C
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