Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - Review - Exercises - Page 578: 18

Answer

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

Work Step by Step

Given $$\int \frac{x^2+8x-3}{x^3+3x^2}\:dx $$ Using partial fractions \begin{aligned} \frac{x^2+8x-3}{x^2\left(x+3\right)}&=\frac{a_0}{x}+\frac{a_1}{x^2}+\frac{a_2}{x+3}\\ &= \frac{a_0x\left(x+3\right)+a_1\left(x+3\right)+a_2x^2}{x^2\left(x+3\right)}\\ x^2+8x-3&=a_0x\left(x+3\right)+a_1\left(x+3\right)+a_2x^2 \end{aligned} at $x= 0\ \ \to \ a_1= -1$, $x= -3 \to a_2 =-2$, then $$x^2+8x-3=a_0x\left(x+3\right)+\left(-1\right)\left(x+3\right)+\left(-2\right)x^2$$ by comparing $$1\cdot \:x^2+8x-3=x^2\left(a_0-2\right)+x\left(3a_0-1\right)-3$$ we get $a_0= 3$ \begin{aligned} \int \frac{x^2+8\:x-3}{x^3+3\:x^2}\:d\:x &=\int \left(\frac{3}{x}-\frac{1}{x^2}-\frac{2}{x+3}\right)dx\\ &= 3\ln \left|x\right|-\left(-\frac{1}{x}\right)-2\ln \left|x+3\right|+C\\ &=3\ln \left|x\right|+\frac{1}{x}-2\ln \left|x+3\right|+C\end{aligned}
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