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 445: 26



Work Step by Step

$$A=\int\frac{s^4+81}{s(s^2+9)^2}ds$$ 1) Express the integrand as a sum of partial fractions: Here we have both an irreducible quadratic factor and a repeated linear factor, $$\frac{s^4+81}{s(s^2+9)^2}=\frac{A}{s}+\frac{Bs+C}{s^2+9}+\frac{Ds+E}{(s^2+9)^2}$$ Clear fractions: $$A(s^4+18s^2+81)+(Bs+C)s(s^2+9)+(Ds+E)s=s^4+81$$ $$As^4+18As^2+81A+Bs^4+9Bs^2+Cs^3+9Cs+Ds^2+Es=s^4+81$$ $$s^4(A+B)+Cs^3+s^2(18A+9B+D)+s(9C+E)+81A=s^4+81$$ Equating coefficients of corresponding powers of $s$, we get $C=0$ and $A=1$ $A+B=1$, so $B=0$ $18A+9B+D=0$, so $D=-18$ $9C+E=0$, so $E=0$ Therefore, $$\frac{s^4+81}{s(s^2+9)^2}=\frac{1}{s}-\frac{18s}{(s^2+9)^2}$$ 2) Evaluate the integral: $$A=\int\frac{1}{s}ds-\int\frac{18s}{(s^2+9)^2}ds$$ $$A=\ln|s|-\int\frac{9}{(s^2+9)^2}d(s^2)$$ $$A=\ln|s|+\frac{9}{s^2+9}+C$$
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