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: 11



Work Step by Step

$$A=\int\frac{x+4}{x^2+5x-6}dx$$ 1) Express the integrand as a sum of partial fractions: $$\frac{x+4}{x^2+5x-6}=\frac{x+4}{(x-1)(x+6)}=\frac{A}{x-1}+\frac{B}{x+6}$$ Clear fractions: $$x+4=Ax+6A+Bx-B$$ $$x+4=(A+B)x+(6A-B)$$ Equating coefficients of corresponding powers of $x$, we get - $A+B=1$ - $6A-B=4$ Calculation gives us $A=5/7$ and $B=2/7$. Therefore, $$\frac{x+4}{x^2+5x-6}=\frac{5}{7(x-1)}+\frac{2}{7(x+6)}$$ 2) Evaluate the integral: $$A=\frac{5}{7}\int\frac{dx}{x-1}+\frac{2}{7}\int\frac{dx}{x+6}$$ $$A=\frac{5}{7}\ln|x-1|+\frac{2}{7}\ln|x+6|+C$$
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