How do you express 1/ (x^4 +1)1x4+1 in partial fractions?

2 Answers
Jun 3, 2016

The polynomial x^4+1x4+1 has not real roots so it is not real factorized then in cannot be expanded in real partial fractions.

Explanation:

The polynomial x^4+1x4+1 has not real roots so it is not real factorized then in cannot be expanded in real partial fractions.

Jun 4, 2016

1/(x^4+1) = (-sqrt(2)x+2)/(x^2-sqrt(2)+1) + (sqrt(2)x+2)/(x^2+sqrt(2)+1)1x4+1=2x+2x22+1+2x+2x2+2+1

Explanation:

x^4+1x4+1 has no linear factors with Real coefficients, since x^4 + 1 >= 1 > 0x4+11>0 for all Real values of xx.

However, it is possible to factor it into quadratic factors:

x^4+1 = (x^2-sqrt(2)+1)(x^2+sqrt(2)x+1)x4+1=(x22+1)(x2+2x+1)

Hence there is a partial fraction decomposition in the form:

1/(x^4+1) = (Ax+B)/(x^2-sqrt(2)+1)+(Cx+D)/(x^2+sqrt(2)x+1)1x4+1=Ax+Bx22+1+Cx+Dx2+2x+1

=((Ax+B)(x^2+sqrt(2)x+1)+(Cx+D)(x^2-sqrt(2)x+1))/(x^4+1)=(Ax+B)(x2+2x+1)+(Cx+D)(x22x+1)x4+1

=((A+C)x^3+(sqrt(2)A+B-sqrt(2)C+D)x^2+(A+sqrt(2)B+C-sqrt(2)D)x+(B+D))/(x^4+1)=(A+C)x3+(2A+B2C+D)x2+(A+2B+C2D)x+(B+D)x4+1

Equating coefficients, we find:

{ (A+C = 0), (sqrt(2)A+B-sqrt(2)C+D=0),(A+sqrt(2)B+C-sqrt(2)D = 0), (B+D=1) :}

Subtracting the first of these from the third, then dividing by sqrt(2), we find:

B-D = 0

Combining this with the fourth equation, we find B=D=1/2

Substituting B+D=1 in the second equation, we deduce:

sqrt(2)A-sqrt(2)C+1 = 0

Combining this with the first equation, we find:

2sqrt(2)A = -1

Hence A=-sqrt(2)/4 and C=sqrt(2)/4

Hence we find:

1/(x^4+1) = (-sqrt(2)x+2)/(x^2-sqrt(2)+1) + (sqrt(2)x+2)/(x^2+sqrt(2)+1)