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: 20



Work Step by Step

$$A=\int\frac{x^2dx}{(x-1)(x^2+2x+1)}=\int\frac{x^2dx}{(x-1)(x+1)^2}$$ 1) Express the integrand as a sum of partial fractions: $$\frac{1}{(x-1)(x+1)^2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$$ Clear fractions: $$A(x+1)^2+B(x^2-1)+C(x-1)=x^2$$ $$Ax^2+2Ax+A+Bx^2-B+Cx-C=x^2$$ $$(A+B)x^2+(2A+C)x+(A-B-C)=x^2$$ Equating coefficients of corresponding powers of $x$, we get - $A+B=1$ - $2A+C=0$ - $A-B-C=0$ Solving for $A$, $B$, $C$, we get $A=1/4$, $B=3/4$ and $C=-1/2$. Therefore, $$\frac{1}{(x-1)(x+1)^2}=\frac{1}{4}\frac{1}{x-1}+\frac{3}{4}\frac{1}{x+1}-\frac{1}{2}\frac{1}{(x+1)^2}$$ 2) Evaluate the integral: $$A=\frac{1}{4}\int\frac{dx}{x-1}+\frac{3}{4}\int\frac{dx}{x+1}-\frac{1}{2}\int\frac{dx}{(x+1)^2}$$ $$A=\frac{1}{4}\ln|x-1|+\frac{3}{4}\ln|x+1|+\frac{1}{2(x+1)}+C$$
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