## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises - Page 549: 9

#### Answer

$$\frac{{1/2}}{{x - 4}} + \frac{{1/2}}{{x + 4}}$$

#### Work Step by Step

\eqalign{ & \frac{{{x^2}}}{{{x^3} - 16x}} \cr & = \frac{{{x^2}}}{{x\left( {{x^2} - 16} \right)}} = \frac{x}{{{x^2} - 16}} \cr & {\text{factoring}} \cr & = \frac{x}{{\left( {x - 4} \right)\left( {x + 4} \right)}} \cr & {\text{partial fraction decomposition}} \cr & \frac{x}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 4}} \cr & x = A\left( {x + 4} \right) + B\left( {x - 4} \right) \cr & {\text{for }}x = 4 \cr & 4 = A\left( {4 + 4} \right) + B\left( {4 - 4} \right) \cr & 4 = 8A \cr & 1/2 = A \cr & {\text{for }}x = - 4 \cr & - 4 = A\left( { - 4 + 4} \right) + B\left( { - 4 - 4} \right) \cr & - 4 = - 8B \cr & 1/2 = B \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 4}} + \frac{B}{{x + 4}} = \frac{{1/2}}{{x - 4}} + \frac{{1/2}}{{x + 4}} \cr}

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