Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 91

Answer

$\frac{1}{2}\ln (9-2x+3x^2)+c.$

Work Step by Step

Let $ u=9-2x+3x^2$ and hence $ du=2(3x-1)dx $, then we have $$ \int \frac{ 3x-1}{9-2x+3x^2}dx=\frac{1}{2}\int \frac{ d u}{u}= \frac{1}{2}\ln |u|+c=\frac{1}{2}\ln |9-2x+3x^2|+c. $$
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