(2x+3)/[(x-2)(x+1)(x2+1)]=A/(x-2)+B/(x+1)+(Cx+D)/(x2+1);
Numerators:
2x+3=A(x+1)(x2+1)+B(x-2)(x2+1)+(Cx+D)(x-2)(x+1);
2x+3=Ax3+Ax2+Ax+A+Bx3-2Bx2+Bx-2B+Cx3-Cx2-2Cx+Dx2-Dx-2D;
Equating coefficients:
x3: A+B+C=0, C=-(A+B)
x2: A-2B-C+D=0=2A-B+D, D=B-2A
x: A+B-2C-D=2=3A+3B-D
From the above two equations: 5A+2B=2
constant: A-2B-2D=3, A-2B-2(B-2A)=3, 5A-4B=3
6B=-1, B=-⅙⇒
5A=3+4B=3-⅔=7/3, A=7/15⇒
C=-(A+B)=-(7/15-1/6)=-3/10⇒
D=B-2A=-11/10.
(2x+3)/[(x-2)(x+1)(x2+1)]=7/(15(x-2))-1/(6(x+1))-(3x+11)/(10(x2+1)).