Precalculus, Chapter 7, 7.4, Section 7.4, Problem 58

(x^3-x+3)/(x^2+x-2)
Divide by applying long division method,
(x^3-x+3)/(x^2+x-2)=(x-1)+(2x+1)/(x^2+x-2)
Now continue with the partial fraction of the remainder expression,
(2x+1)/(x^2+x-2)=(2x+1)/(x^2-x+2x-2)
=(2x+1)/(x(x-1)+2(x-1))
=(2x+1)/((x-1)(x+2))
Let (2x+1)/(x^2+x-2)=A/(x-1)+B/(x+2)
(2x+1)/(x^2+x-2)=(A(x+2)+B(x-1))/((x-1)(x+2))
(2x+1)/(x^2+x-2)=(Ax+2A+Bx-B)/((x-1)(x+2))
:.(2x+1)=Ax+2A+Bx-B
2x+1=x(A+B)+2A-B
Equating the coefficients of the like terms,
A+B=2 ---- equation 1
2A-B=1 --- equation 2
Add the equation 1 and 2,
A+2A=2+1
3A=3
A=1
Plug the value of A in the equation 1,
1+B=2
B=2-1
B=1
:.(x^3-x+3)/(x^2+x-2)=x-1+1/(x-1)+1/(x+2)