Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.5 The Method of Partial Fractions - Exercises - Page 423: 2


$$A=-3$$ $$B=5$$

Work Step by Step

$\frac{2x-3}{(x-3)(x-4)}$ = $\frac{A}{x-3}$ + $\frac{B}{x-4}$ Multiplying (x-3)(x-4) on both sides, we have: $2x-3=A(x-4)+B(x-3)$ If we let x=4 on both sides, we can eliminate constant A to find constant B: $8-3=A(4-4)+B(4-3)$ Thus, $B = 5$. Similarly, if we let x = 3 on both sides, we can eliminate constant B to find constant A: $6-3=A(3-4)+B(3-3)$ Thus, $3=-A$ and so $A = -3$.
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