x2+x+4(1−x)(x2+1)=A1−x+Bx+Cx2+1
= A(x2+1)+(Bx+C)(1−x)(1−x)(x2+1)
impliesx2+x+4=A(x2+1)+(Bx+C)(1−x)
x2+x+4=Ax2+A+Bx−Bx2−Cx+C
implies(A−B)x2=x2;A−B=1......(i)
(B−C)x=x;B−C=1.....(ii)
A+C=4......(iii)
Solving the above simultaneous equations by any of the known methods, we get
A=3,B=2,C=1
∴x2+x+4(1−x)(x2+1)=31−x+2x+1x2+1