Question

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

Answer 1

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

Answer 2

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

Explanation:

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

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

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

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

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)`