# How do you express (x^4)/((x²+1)(x²-1)) in partial fractions?

##### 1 Answer
Nov 30, 2016

The answer is $= 1 + \frac{- \frac{1}{2}}{{x}^{2} + 1} + \frac{- \frac{1}{4}}{x + 1} + \frac{\frac{1}{4}}{x - 1}$

#### Explanation:

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

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

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

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

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

1=(Ax+B)(x^2-1)+C(x-1)(x^2+1)+D(x+1)(x^2+1))

Let $x = 1$, $\implies$, $1 = 4 D$, $\implies$, $D = \frac{1}{4}$

Let $x = - 1$, $\implies$, $1 = - 4 C$, $\implies$, $C = - \frac{1}{4}$

No coefficients, $1 = - B - C + D$, $\implies$, $B = \frac{1}{2} - 1 = - \frac{1}{2}$

Coefficients of ${x}^{3}$, $\implies$, $0 = A + C + D$, $\implies$. $A = 0$

Therefore,

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