Calculus (3rd Edition)

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



Work Step by Step

By using the fact that $\int u'/u = \ln u$, we have $$\int \frac{3x-1}{9-2x+3x^2}dx=\frac{1}{2}\int \frac{6x-2}{9-2x+3x^2}dx=\frac{1}{2}\ln(9-2x+3x^2)+c.$$
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