(3x+2)/(x-3)(x^2+2)=A/(x-3)+(Bx+C)/(x^2+2), so Ax^2+2A+Bx^2+Cx-3Bx-3C=3x+2.
Equate x^2: A+B=0 (no x^2 term), so B=-A;
Equate x: C-3B=3, so C=3+3B, C=3-3A by substitution;
Equate constant: 2A-3C=2; 2A-3(3-3A)=2; 11A-9=2; 11A=11, so A=1, B=-1, C=0.
Partial fractions are: 1/(x-3)-x/(x^2+2)
Check: (x^2+2-x^2+3x)/(x-3)(x^2+2)=(3x+2)/((x-3)(x^2+2) Right!