The correct option is C 2
3x2+1(x2+1)=Ax+Bx2+1+Cx+D(x2+1)2+ex+f(x2+1)3+9x+4(x2+1)4
LHS=RHS
3x2+1=(Ax+B)[x2+1]3+(Cx+D)(x2+1)2
+(ex+f)(x2+1)
+(9x+4)
comparing co-efficients of
x7,x6,x5,x4,x3 in LHS & RHS
A=B=C=D=e=o
So f(x2+1)+(9x+4) is the Use
3x2+1(x2+1)4=f(x2+1)3+9x+4(x2+1)4
So, two partial fractions.