University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 446: 35



Work Step by Step

$$I=\int\frac{9x^3-3x+1}{x^3-x^2}dx$$ 1) Perform long division: We perform long division to rewrite the fraction as a polynomial plus a proper fraction, which is $$\frac{9x^3-3x+1}{x^3-x^2}=9+\frac{9x^2-3x+1}{x^3-x^2}=9+\frac{9x^2-3x+1}{x^2(x-1)}$$ 2) Express the remainder fraction as a sum of partial fractions: $$\frac{9x^2-3x+1}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}$$ Clear fractions: $$Ax(x-1)+B(x-1)+Cx^2=9x^2-3x+1$$ $$Ax^2-Ax+Bx-B+Cx^2=9x^2-3x+1$$ $$(A+C)x^2+(-A+B)x-B=9x^2-3x+1$$ Equating coefficients of corresponding powers of $x$, we get $-B=1$, so $B=-1$ $-A+B=-3$, so $A=2$ $A+C=9$, so $C=7$ Therefore, $$\frac{9x^2-3x+1}{x^2(x-1)}=\frac{2}{x}-\frac{1}{x^2}+\frac{7}{x-1}$$ 2) Evaluate the integral: $$I=\int9dx+2\int\frac{1}{x}dx-\int\frac{1}{x^2}dx+7\int\frac{1}{x-1}dx$$ $$I=9x+2\ln|x|+\frac{1}{x}+7\ln|x-1|+C$$
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