Since numerator and denominator have the same degree, include a constant term: \[\frac{4x^2-13x+13}{(2x-1)(x-3)}=A+\frac{B}{2x-1}+\frac{C}{x-3}.\] Multiply by the denominator: \[4x^2-13x+13=A(2x^2-7x+3)+B(x-3)+C(2x-1).\] Comparing quadratic coefficients gives A=2. At x=1/2, the left side is 15/2 and the right side is −5B/2, giving B=−3. At x=3, the left side is 10 and the right side is 5C, giving C=2. Thus \[\boxed{2-\frac3{2x-1}+\frac2{x-3}},\qquad x\ne\tfrac12,3.\]