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
$$
\int \frac{ 3x-1}{9-2x+3x^2}dx=\frac{1}{2}\int \frac{ d u}{u}= \frac{1}{2}\ln |u|+c=\frac{1}{2}\ln |9-2x+3x^2|+c.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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