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: 10



Work Step by Step

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