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: 10

Answer

$$\frac{{1/5}}{{x - 4}} + \frac{{4/5}}{{x + 1}}$$

Work Step by Step

$$\eqalign{ & \frac{{{x^2} - 3x}}{{{x^3} - 3{x^2} - 4x}} \cr & {\text{factoring}} \cr & = \frac{{x\left( {x - 3} \right)}}{{x\left( {{x^2} - 3x - 4} \right)}} = \frac{{x - 3}}{{{x^2} - 3x - 4}} \cr & = \frac{{x - 3}}{{\left( {x - 4} \right)\left( {x + 1} \right)}} \cr & {\text{partial fraction decomposition}} \cr & \frac{{x - 3}}{{\left( {x - 4} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 1}} \cr & x - 3 = A\left( {x + 1} \right) + B\left( {x - 4} \right) \cr & {\text{for }}x = 4 \cr & 4 - 3 = A\left( {4 + 1} \right) + B\left( {4 - 4} \right) \cr & 1 = 5A \cr & 1/5 = A \cr & {\text{for }}x = - 1 \cr & - 1 - 3 = A\left( { - 1 + 1} \right) + B\left( { - 1 - 4} \right) \cr & - 4 = - 5B \cr & 4/5 = B \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 4}} + \frac{B}{{x + 1}} = \frac{{1/5}}{{x - 4}} + \frac{{4/5}}{{x + 1}} \cr} $$
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