# How do you express (x^4+6)/(x^5+7x^3) in partial fractions?

##### 1 Answer
Oct 31, 2016

The result is $= \frac{- \frac{6}{49}}{x} + \frac{\frac{6}{7}}{x} ^ 3 + \frac{\frac{55}{49} x}{{x}^{2} + 7}$

#### Explanation:

Let's factorise the denominator ${x}^{5} + 7 {x}^{3} = {x}^{3} \left({x}^{2} + 7\right)$
So $\frac{{x}^{4} + 6}{{x}^{3} \left({x}^{2} + 7\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x} ^ 3 + \frac{D x + E}{{x}^{2} + 7}$
$= \frac{A {x}^{2} \left({x}^{2} + 7\right) + B x \left({x}^{2} + 7\right) + C \left({x}^{2} + 7\right) + \left(D x + E\right) \left({x}^{2} + 7\right)}{{x}^{3} \left({x}^{2} + 7\right)}$
${x}^{4} + 6 = A {x}^{2} \left({x}^{2} + 7\right) + B x \left({x}^{2} + 7\right) + C \left({x}^{2} + 7\right) + \left(D x + E\right) {x}^{3}$
$x = 0$$\implies$$6 = 7 C$$\implies$$C = \frac{6}{7}$
coefficients ${x}^{4}$$\implies$$1 = A + D$
coefficients ${x}^{3}$$\implies$$0 = B + E$
coefficients ${x}^{2}$$\implies$$0 = 7 A + C$
coefficients of $x$$\implies$$0 = 7 B$
$E = 0$
$A = - \frac{C}{7} = - \frac{6}{49}$
$B = 0$
$D = 1 - A = 1 + \frac{6}{49} = \frac{55}{49}$
Finally we have
$\frac{{x}^{4} + 6}{{x}^{3} \left({x}^{2} + 7\right)} = \frac{- \frac{6}{49}}{x} + \frac{\frac{6}{7}}{x} ^ 3 + \frac{\frac{55}{49} x}{{x}^{2} + 7}$