⇒9x−7=A(x2+1)+(Bx+C)(x+3) ⇒9x−7=Ax2+A+Bx2+3Bx+Cx+3C Comparing coefficients of x2,x and constant to both the sides, we have A+B=0 . . . (i) 3B+C=9 . . . (ii) A+3C=−7... (iii) From Eq. (i), A=−B ∴ Eq. (ii) becomes, −3A+C=9 . . . (iv) Solving Eqs. (iii) and (iv), we have C=