## University Calculus: Early Transcendentals (3rd Edition)

$\frac{2}{x-3}+\frac{3}{x-2}$
The form of the partial fraction decomposition is $\frac{5x-13}{(x-3)(x-2)}=\frac{A}{x-3}+\frac{B}{x-2}$ $⇒5x-13= A(x-2)+B(x-3)$ Equating the coefficients of x and the constant term, we get A+B=5 and 2A+3B=13. Solving these equations, we get A=2 and B=3. Thus, $\frac{5x-13}{(x-3)(x-2)}=\frac{2}{x-3}+\frac{3}{x-2}$