⇒x2+3=(x2+2)(Ax−B)+(x2+1)(Cx+D) ⇒x2+3=Ax3+Bx2+2Ax+2B+Cx3+x2D+Cx+D ⇒x2+3=(A+C)x3+(B+D)x2+(2A+C)x+2B+D On comparing, A+C=0⋅⋅⋅⋅⋅⋅⋅(i) B+D=1⋅⋅⋅⋅⋅⋅⋅(ii) 2A+C=0⋅⋅⋅⋅⋅⋅⋅(iii) 2B+D=3⋅⋅⋅⋅⋅⋅⋅(iv) From Eqs. (i) and (iii), A=C=0 From Eqs. (ii) and (iv), B=2 and D=−1 ∴A+B+C+D=1