Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 501: 18

Answer

$$\int_{1}^{2}\frac{3x^{2}+6x+2}{x^{2}+3x+2}dx=3+ln\frac{3}{8}$$

Work Step by Step

$$\int_{1}^{2}\frac{3x^{2}+6x+2}{x^{2}+3x+2}dx=\int_{1}^{2}(3+\frac{-3x-4}{x^{2}+3x+2})dx$$ $$=\int_{1}^{2}(3+\frac{A}{x+2}+\frac{B}{x+1})dx$$ $3x^{2}+(A+B+9)x+(A+2B+6)=3x^{2}+6x+2. So\ A=-2, B=-1$ $$\int_{1}^{2}\frac{3x^{2}+6x+2}{x^{2}+3x+2}dx=\int_{1}^{2}(3-\frac{2}{x+2}-\frac{1}{x+1})dx$$ $$=\left [ 3x-2ln|x+2|-ln|x+1| \right ]_{1}^{2}$$ $$=6-2ln4-ln3-3+2ln3+ln2=3+ln\frac{3}{8}$$
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