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

Jun 3, 2016

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.

Jun 4, 2016

$\frac{1}{{x}^{4} + 1} = \frac{- \sqrt{2} x + 2}{{x}^{2} - \sqrt{2} + 1} + \frac{\sqrt{2} x + 2}{{x}^{2} + \sqrt{2} + 1}$

#### Explanation:

${x}^{4} + 1$ has no linear factors with Real coefficients, since ${x}^{4} + 1 \ge 1 > 0$ for all Real values of $x$.

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

${x}^{4} + 1 = \left({x}^{2} - \sqrt{2} + 1\right) \left({x}^{2} + \sqrt{2} x + 1\right)$

Hence there is a partial fraction decomposition in the form:

$\frac{1}{{x}^{4} + 1} = \frac{A x + B}{{x}^{2} - \sqrt{2} + 1} + \frac{C x + D}{{x}^{2} + \sqrt{2} x + 1}$

$= \frac{\left(A x + B\right) \left({x}^{2} + \sqrt{2} x + 1\right) + \left(C x + D\right) \left({x}^{2} - \sqrt{2} x + 1\right)}{{x}^{4} + 1}$

$= \frac{\left(A + C\right) {x}^{3} + \left(\sqrt{2} A + B - \sqrt{2} C + D\right) {x}^{2} + \left(A + \sqrt{2} B + C - \sqrt{2} D\right) x + \left(B + D\right)}{{x}^{4} + 1}$

Equating coefficients, we find:

$\left\{\begin{matrix}A + C = 0 \\ \sqrt{2} A + B - \sqrt{2} C + D = 0 \\ A + \sqrt{2} B + C - \sqrt{2} D = 0 \\ B + D = 1\end{matrix}\right.$

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 $B = D = \frac{1}{2}$

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:

$2 \sqrt{2} A = - 1$

Hence $A = - \frac{\sqrt{2}}{4}$ and $C = \frac{\sqrt{2}}{4}$

Hence we find:

$\frac{1}{{x}^{4} + 1} = \frac{- \sqrt{2} x + 2}{{x}^{2} - \sqrt{2} + 1} + \frac{\sqrt{2} x + 2}{{x}^{2} + \sqrt{2} + 1}$