Let the expression equal A/(x+1)+B/(x-2)+C/(x+3).
A(x^2+x-6)+B(x^2+4x+3)+C(x^2-x-2)=x^2+3x-2
1) Equate x^2: A+B+C=1
2) Equate x: A+4B-C=3
3) Equate constant: -6A+3B-2C=-2
4) (1)+(2): 2A+5B=4
5) 2*(1)+(3): -4A+5B=0
6) 2*(4)+(5): 15B=8, so B=8/15; 2A+8/3=4, so A=2/3 and C=1-2/3-8/15=-1/5
So the partial fractions are:
2/(3(x+1))+8/(15(x-2))-1/(5(x+3))