## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

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

#### Answer

$$\frac{3}{{x - 2}} + \frac{2}{{x - 1}}$$

#### Work Step by Step

\eqalign{ & \frac{{5x - 7}}{{{x^2} - 3x + 2}} \cr & {\text{factoring}} \cr & = \frac{{5x - 7}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} \cr & {\text{partial fraction decomposition}} \cr & \frac{{5x - 7}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \frac{A}{{x - 2}} + \frac{B}{{x - 1}} \cr & 5x - 7 = A\left( {x - 1} \right) + B\left( {x - 2} \right) \cr & {\text{letting }}x = 2 \cr & 5\left( 2 \right) - 7 = A\left( {2 - 1} \right) + B\left( {2 - 2} \right) \cr & 3 = A \cr & {\text{letting }}x = 1 \cr & 5\left( 1 \right) - 7 = A\left( {1 - 1} \right) + B\left( {1 - 2} \right) \cr & - 2 = B\left( { - 1} \right) \cr & B = 2 \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 2}} + \frac{B}{{x - 1}} = \frac{3}{{x - 2}} + \frac{2}{{x - 1}} \cr}

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