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: 16



Work Step by Step

$$A=\int\frac{x+3}{2x^3-8x}dx$$ 1) Express the integrand as a sum of partial fractions: $$\frac{x+3}{2x^3-8x}=\frac{x+3}{2x(x^2-4)}=\frac{x+3}{2x(x-2)(x+2)}$$ $$=\frac{A}{2x}+\frac{B}{x-2}+\frac{C}{x+2}$$ Clear fractions: $$x+3=A(x^2-4)+B(2x^2+4x)+C(2x^2-4x)$$ $$x+3=Ax^2-4A+2Bx^2+4Bx+2Cx^2-4Cx$$ $$x+3=(A+2B+2C)x^2+(4B-4C)x-4A$$ Equating coefficients of corresponding powers of $x$, we get - $-4A=3$ - $A+2B+2C=0$ - $4B-4C=1$ Calculation gives us $A=-3/4$, $B=5/16$ and $C=1/16$ Therefore, $$\frac{x+3}{2x^3-8x}=\frac{5}{16(x-2)}+\frac{1}{16(x+2)}-\frac{3}{8x}$$ 2) Evaluate the integral: $$A=\frac{5}{16}\int\frac{dx}{x-2}+\frac{1}{16}\int\frac{dx}{x+2}-\frac{3}{8}\int\frac{dx}{x}$$ $$A=\frac{5}{16}\ln|x-2|+\frac{1}{16}\ln|x+2|-\frac{3}{8}\ln|x|+C$$
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