Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 515: 8

Answer

$3\ln \left| x \right| - 2\ln \left| {x - 4} \right| + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{x - 12}}{{{x^2} - 4x}}} dx \cr & {\text{Factoring the integrand}} \cr & = \int {\frac{{x - 12}}{{x\left( {x - 4} \right)}}} dx \cr & {\text{Using the method of partial fractions}} \cr & \frac{{x - 12}}{{x\left( {x - 4} \right)}} = \frac{A}{x} + \frac{B}{{x - 4}} \cr & {\text{Multiply the equation by }}x\left( {x - 4} \right) \cr & x - 12 = A\left( {x - 4} \right) + Bx{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Let }}x = 0{\text{ into }}\left( {\bf{1}} \right) \cr & 0 - 12 = A\left( {0 - 4} \right) + B\left( 0 \right){\text{ }} \to A = 3 \cr & {\text{Let }}x = 4{\text{ into }}\left( {\bf{1}} \right) \cr & 4 - 12 = A\left( {4 - 4} \right) + B\left( 4 \right){\text{ }} \to B = - 2 \cr & {\text{Substituting }}A{\text{ and }}B{\text{ into the partial fraction decomposition}} \cr & \frac{{x - 12}}{{x\left( {x - 4} \right)}} = \frac{3}{x} - \frac{2}{{x - 4}} \cr & {\text{Therefore}}{\text{,}} \cr & \int {\frac{{x - 12}}{{{x^2} - 4x}}} dx = \int {\left( {\frac{3}{x} - \frac{2}{{x - 4}}} \right)} dx \cr & {\text{Integrating}} \cr & = 3\ln \left| x \right| - 2\ln \left| {x - 4} \right| + C \cr} $$
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