Calculus (3rd Edition)

$$A=-3$$ $$B=5$$
$\frac{2x-3}{(x-3)(x-4)}$ = $\frac{A}{x-3}$ + $\frac{B}{x-4}$ 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 $A = -3$.