# How do you write the partial fraction decomposition of the rational expression  1 / ((x^2 + 1) (x^2 +4))?

Dec 16, 2015

Solve to find:

$\frac{1}{\left({x}^{2} + 1\right) \left({x}^{2} + 4\right)} = \frac{1}{3 \left({x}^{2} + 1\right)} - \frac{1}{3 \left({x}^{2} + 4\right)}$

#### Explanation:

Neither of the quadratics $\left({x}^{2} + 1\right)$ and $\left({x}^{2} + 4\right)$ have linear factors with Real coefficients, so let's leave them as quadratics and attempt to solve:

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

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

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

Equating coefficients we find:

$A + B = 0$

$4 A + B = 1$

Hence $A = \frac{1}{3}$ and $B = - \frac{1}{3}$

So:

$\frac{1}{\left({x}^{2} + 1\right) \left({x}^{2} + 4\right)} = \frac{1}{3 \left({x}^{2} + 1\right)} - \frac{1}{3 \left({x}^{2} + 4\right)}$

If we allow Complex coefficients, then we find:

$\frac{1}{{x}^{2} + 1} = \frac{i}{2 \left(x + i\right)} - \frac{i}{2 \left(x - i\right)}$

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

Hence:

$\frac{1}{3 \left({x}^{2} + 1\right)} - \frac{1}{3 \left({x}^{2} + 4\right)}$

$= \frac{i}{6 \left(x + i\right)} - \frac{i}{6 \left(x - i\right)} + \frac{i}{12 \left(x - 2 i\right)} - \frac{i}{12 \left(x + 2 i\right)}$