# How do you express (x^5 + 1)/(x^6 - x^4) in partial fractions?

##### 1 Answer
Mar 8, 2016

$\frac{{x}^{5} + 1}{{x}^{4} \left(x + 1\right) \left(x - 1\right)} = - \frac{1}{x} ^ 2 - \frac{1}{x} ^ 4 + \frac{1}{x - 1}$

#### Explanation:

$\frac{{x}^{5} + 1}{{x}^{4} \left({x}^{2} - 1\right)} = \frac{{x}^{5} + 1}{{x}^{4} \left(x + 1\right) \left(x - 1\right)}$
$\frac{{x}^{5} + 1}{{x}^{4} \left(x + 1\right) \left(x - 1\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x} ^ 3 + \frac{D}{x} ^ 4 + \frac{E}{x + 1} + \frac{F}{x - 1}$

${x}^{5} + 1 = A \left({x}^{3}\right) \left({x}^{2} - 1\right) + B \left({x}^{2}\right) \left({x}^{2} - 1\right) + C \left(x\right) \left({x}^{2} - 1\right) + D \left({x}^{2} - 1\right) + E \left({x}^{4}\right) \left(x - 1\right) + F \left({x}^{4}\right) \left(x + 1\right)$

${x}^{5} + 1 = A {x}^{5} - A {x}^{3} + B {x}^{4} - B {x}^{2} + C {x}^{3} - C x + D {x}^{2} - D + E {x}^{5} - E {x}^{4} + F {x}^{5} + F {x}^{4}$

$1 = A + E + F , 0 = B - E + F , 0 = - A + C , 0 = - B + D , 0 = - C , 1 = - D$

$A = 0 , B = - 1 , C = 0 , D = - 1 , E = 0 , F = 1$

$\frac{{x}^{5} + 1}{{x}^{4} \left(x + 1\right) \left(x - 1\right)} = - \frac{1}{x} ^ 2 - \frac{1}{x} ^ 4 + \frac{1}{x - 1}$