# How do you express (x^4+1)/(x^5+6 x^3) in partial fractions?

Sep 15, 2016

$\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}} = - \frac{1}{36 x} + \frac{1}{6 {x}^{3}} + \frac{37 x}{36 \left({x}^{2} + 6\right)}$

#### Explanation:

$\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}} = \frac{{x}^{4} + 1}{{x}^{3} \left({x}^{2} + 6\right)}$

$\textcolor{w h i t e}{\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}}} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x} ^ 3 + \frac{D x + E}{{x}^{2} + 6}$

$\textcolor{w h i t e}{\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}}} = \frac{A {x}^{2} \left({x}^{2} + 6\right) + B x \left({x}^{2} + 6\right) + C \left({x}^{2} + 6\right) + D {x}^{4} + E {x}^{3}}{{x}^{2} + 6}$

$\textcolor{w h i t e}{\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}}} = \frac{\left(A + D\right) {x}^{4} + \left(B + E\right) {x}^{3} + \left(6 A + C\right) {x}^{2} + 6 B x + 6 C}{{x}^{2} + 6}$

Equating coefficients gives us this system of linear equations:

$\left\{\begin{matrix}A + D = 1 \\ B + E = 0 \\ 6 A + C = 0 \\ 6 B = 0 \\ 6 C = 1\end{matrix}\right.$

Hence:

$\left\{\begin{matrix}A = - \frac{1}{36} \\ B = 0 \\ C = \frac{1}{6} \\ D = \frac{37}{36} \\ E = 0\end{matrix}\right.$

So:

$\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}} = - \frac{1}{36 x} + \frac{1}{6 {x}^{3}} + \frac{37 x}{36 \left({x}^{2} + 6\right)}$