Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Review - Exercises - Page 538: 18


$ 3\displaystyle \ln|x|+\frac{1}{x}-2\ln|x+3|+C$

Work Step by Step

$x^{3}+3x^{2}=x^{2}(x+3)$ $\displaystyle \frac{x^{2}+8x-3}{x^{3}+3x^{2}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+3}$ $\displaystyle \frac{x^{2}+8x-3}{x^{3}+3x^{2}}=\frac{Ax(x+3)+B(x+3)+Cx^{2}}{x^{2}(x+3)}$ $x^{2}+8x-3=(A+C)x^{2}+(3A+B)x+3B\Rightarrow\left\{\begin{array}{ll} 3B=-3 & \Rightarrow B=-1\\ 3A+B=8 & \Rightarrow A=3\\ A+C=1 & \Rightarrow C=-2 \end{array}\right.$ $\displaystyle \int\frac{x^{2}+8x-3}{x^{3}+3x^{2}}dx=\int\left(\frac{3}{x}+\frac{-1}{x^{2}}+\frac{-2}{x+3}\right)dx$ $ = 3\displaystyle \ln|x|-\frac{x^{-1}}{-1}-2\ln|x+3|+C$ $ = 3\displaystyle \ln|x|+\frac{1}{x}-2\ln|x+3|+C$
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