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: 9



Work Step by Step

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