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

$\frac{3}{x-2}+\frac{2}{x-1}$
$\frac{5x-7}{x^{2}-3x+2}=\frac{5x-7}{(x-2)(x-1)}$ Decomposing to partial fractions, we have $\frac{5x-7}{(x-2)(x-1)}=\frac{A}{x-2}+\frac{B}{x-1}$ This gives $5x-7= A(x-1)+B(x-2)$. Equating the coefficients of x and the constant term, we obtain A+B=5 and A+2B=7. Solving these equations, we get A=3 and B=2. Thus $\frac{5x-7}{x^{2}-3x+2}=\frac{3}{x-2}+\frac{2}{x-1}$