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: 15



Work Step by Step

$$A=\int\frac{dt}{t^3+t^2-2t}$$ 1) Express the integrand as a sum of partial fractions: $$\frac{1}{t^3+t^2-2t}=\frac{1}{t(t^2+t-2)}=\frac{1}{t(t-1)(t+2)}$$ $$=\frac{A}{t}+\frac{B}{t-1}+\frac{C}{t+2}$$ Clear fractions: $$1=A(t^2+t-2)+B(t^2+2t)+C(t^2-t)$$ $$1=At^2+At-2A+Bt^2+2Bt+Ct^2-Ct$$ $$1=(A+B+C)t^2+(A+2B-C)t-2A$$ Equating coefficients of corresponding powers of $t$, we get - $-2A=1$ - $A+B+C=0$ - $A+2B-C=0$ Calculation gives us $A=-1/2$, $B=1/3$ and $C=1/6$ Therefore, $$\frac{1}{t^3+t^2-2t}=\frac{1}{3(t-1)}+\frac{1}{6(t+2)}-\frac{1}{2t}$$ 2) Evaluate the integral: $$A=\frac{1}{3}\int\frac{dt}{t-1}+\frac{1}{6}\int\frac{dt}{t+2}-\frac{1}{2}\int\frac{dt}{t}$$ $$A=\frac{1}{3}\ln|t-1|+\frac{1}{6}\ln|t+2|-\frac{1}{2}\ln|t|+C$$
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