Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises: 5

Answer

$$\frac{1}{{3\left( {x - 4} \right)}} - \frac{1}{{3\left( {x + 2} \right)}}$$

Work Step by Step

$$\eqalign{ & \frac{2}{{{x^2} - 2x - 8}} \cr & {\text{factoring}} \cr & = \frac{2}{{\left( {x - 4} \right)\left( {x + 2} \right)}} \cr & {\text{partial fraction decomposition}} \cr & \frac{2}{{\left( {x - 4} \right)\left( {x + 2} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 2}} \cr & 2 = A\left( {x + 2} \right) + B\left( {x - 4} \right) \cr & {\text{letting }}x = 4 \cr & 2 = A\left( {4 + 2} \right) + B\left( {4 - 4} \right) \cr & 2 = A\left( 6 \right) \cr & A = 1/3 \cr & {\text{letting }}x = - 2 \cr & 2 = A\left( { - 2 + 2} \right) + B\left( { - 2 - 4} \right) \cr & 2 = B\left( { - 6} \right) \cr & B = - 1/3 \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 4}} + \frac{B}{{x + 2}} = \frac{{1/3}}{{x - 4}} - \frac{{1/3}}{{x + 2}} \cr & = \frac{1}{{3\left( {x - 4} \right)}} - \frac{1}{{3\left( {x + 2} \right)}} \cr} $$
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