# How do you write the partial fraction decomposition of the rational expression  (x^4+1)/(x^5+6 x^3)?

Oct 11, 2016

(x^4+1)/(x^5+6 x^3)=1/(6 x^3) - 1/(36 x) + 37/(72 ( x-i sqrt[6] )) + 37/( 72 (x+i sqrt[6]))

#### Explanation:

The proposition
$\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}} = \frac{{x}^{4} + 1}{{x}^{3} \left({x}^{2} + 6\right)} = {c}_{1} / x + {c}_{2} / {x}^{2} + {c}_{3} / {x}^{3} + {c}_{4} / \left(x + i \sqrt{6}\right) + {c}_{5} / \left(x - i \sqrt{6}\right)$
The ${c}_{k}$ can be determined according to some techniques. The most elementar is

$\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}} = \frac{1 - 6 {c}_{3} - 6 {c}_{2} x - \left(6 {c}_{1} + {c}_{3}\right) {x}^{2} + \left(i \sqrt{6} {c}_{4} - i \sqrt{6} {c}_{5} - {c}_{2}\right) {x}^{3} + \left(1 - {c}_{1} - {c}_{4} - {c}_{5}\right) {x}^{4}}{{x}^{3} \left(6 + {x}^{2}\right)}$

2) Choosing ${c}_{k}$ such that
${x}^{4} + 1 = \left(1 - 6 {c}_{3} - 6 {c}_{2} x - \left(6 {c}_{1} + {c}_{3}\right) {x}^{2} + \left(i \sqrt{6} {c}_{4} - i \sqrt{6} {c}_{5} - {c}_{2}\right) {x}^{3} + \left(1 - {c}_{1} - {c}_{4} - {c}_{5}\right) {x}^{4}\right) , \forall x \in \mathbb{R}$

3) Solving the conditions

{(1 - 6 c_3=0), (-6 c_2=0), (6 c_1 + c_3=0), (-c_2 + i sqrt[6] c_4 - i sqrt[6] c_5=0), (1 - c_1 - c_4 - c_5=0):}

obtaining

${c}_{1} = - \frac{1}{36} , {c}_{2} = 0 , {c}_{3} = \frac{1}{6} , {c}_{4} = \frac{37}{72} , {c}_{5} = \frac{37}{72}$

(x^4+1)/(x^5+6 x^3)=1/(6 x^3) - 1/(36 x) + 37/(72 ( x-i sqrt[6] )) + 37/( 72 (x+i sqrt[6]))

Note. We can avoid the complex expansion doing

$\frac{c {'}_{3} x + c {'}_{4}}{{x}^{2} + 6}$ instead of ${c}_{4} / \left(x + i \sqrt{6}\right) + {c}_{5} / \left(x - i \sqrt{6}\right)$