How do you express $\frac{2 x + 3}{x^{4} - 9 x^{2}}$ $(2x+ 3)/(x^4-9x^2)$ in partial fractions?

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How do you express $\frac{2 x + 3}{x^{4} - 9 x^{2}}$ $(2x+ 3)/(x^4-9x^2)$ in partial fractions?

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Answer 1 · 2016-05-13T16:41:07

$\frac{2 x + 3}{x^{4} - 9 x^{2}} = - \frac{2}{9 x} + \frac{1}{3 x^{2}} - \frac{1}{18 (x + 3)} - \frac{1}{6 (x - 3)}$ $(2x+3)/(x^4-9x^2)=-2/(9x)+1/(3x^2)-1/(18(x+3))-1/(6(x-3))$

Explanation:

Looking to the denominator $x^{4} - 9 x^{2} = x^{2} (x + 3) (x - 3)$ $x^4-9x^2=x^2(x+3)(x-3)$ then the fraction admits an expansion such that
$\frac{2 x + 3}{x^{4} - 9 x^{2}} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x + 3} + \frac{D}{x - 3}$ $(2x+3)/(x^4-9x^2)=A/x+B/x^2+C/(x+3)+D/(x-3)$ so
doing $f = \frac{2 x + 3}{x^{4} - 9 x^{2}} - \frac{A}{x} - \frac{B}{x^{2}} - \frac{C}{x + 3} - \frac{D}{x - 3} = 0$ $f = (2x+3)/(x^4-9x^2)-A/x-B/x^2-C/(x+3)-D/(x-3)=0$ and calculanting $f (x^{4} - 9 x^{2}) = (D + C - A) x^{3} + (3 D + B - 3 C) x^{2} + (9 A + 2) x + 3 - 9 B = 0$ $f(x^4-9x^2) = (D+C-A)x^3+(3D+B-3C)x^2+(9A+2)x+3-9B =0$ for all values of x, we get:
$A = - \frac{2}{9}, B = \frac{1}{3}, C = - \frac{1}{18}, D = - \frac{1}{6}$ $A=-2/9,B=1/3,C=-1/18,D=-1/6$

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How do you express $\frac{2 x + 3}{x^{4} - 9 x^{2}}$ $(2x+ 3)/(x^4-9x^2)$ in partial fractions?

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