Start with: # (6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )#
The denominator is already factored into irreducible polynomials (over the Reals).
Because #x^2+1# cannnot be factored using Real number coefficients, we need a linear numerator for one of the fractions
We need:
#A/(4x+1)+(Bx+C)/(x^2+1) = (6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )#
This lead to:
#(Ax^2+A+4Bx^2+4Cx+Bx+C) / ( (4x+1)(x^2 +1) )=(6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )#
And so:
#((A+4B)x^2+(B+4C)x+(A+C)) / ( (4x+1)(x^2 +1) )=(6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )#
So we need to solve the system:
#A# #+4B# #" " " " # # =# #6#
#" " " " # #B# #+4C# # =# #-3#
#A# #" " " " " # # +C# #=# #1#
Eq3 implies #A = 1-C# and substituting in Eq1 and simplifying gets us:
#" " " " # #4B# #-C# # =# #5#
Eq2 is
#" " " " # #B# #+4C# # =# #-3#
So
#" " " "# #-4B# #-16C# # =# #12#
Thus #-17C = 17# and #C= -1#
Knowing #C#, we can find #A = 1-(-1) =2#
And #B+4(-1) = -3# gets us #B = 1#
#A/(4x+1)+(Bx+C)/(x^2+1) = 2/(4x+1)+(x-1)/(x^2+1)#
