Answer
$$\frac{1}{2}\ln(9-2x+3x^2)+c.$$
Work Step by Step
By using the fact that $\int u'/u = \ln u$, we have
$$\int \frac{3x-1}{9-2x+3x^2}dx=\frac{1}{2}\int \frac{6x-2}{9-2x+3x^2}dx=\frac{1}{2}\ln(9-2x+3x^2)+c.$$
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 - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 100
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