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I have a partial fraction problem where I need to decompose $$\frac{1}{(-2x-x^4)}$$ which becomes $$\frac{1}{(x)(-x^3-2)}.$$

I'm used to dealing with partial fractions where the factor $(x-2)$ is raised to the third power, like $(x-2)^3$, but what do I do when the third power is inside the factor, like $(x^3-2)$?

It no longer seems to fit any of the forms for partial fraction decomposition, and yet, apparently it can still be done. How?

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You need a numerator with one less power of $x$ than the term in the denominator, so you would write $\frac {-1}{x^4+2x}=\frac ax + \frac {bx^2+cx+d}{x^3+2}$ and find $a,b,c,d$ If we multiply by $x$ and take the limit as $x \to 0$ we find $a=-\frac 12$. If we clear the fractions we find $-1=-\frac 12(x^3+2)+bx^3+cx^2+dx$ which gives $b=\frac 12,c=0,d=0$ and $$\frac {-1}{x^4+2x}=-\frac 1{2x} + \frac {x^2}{2(x^3+2)}$$ Alpha confirms this.

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If so you need to write before $$x^3+2=(x+\sqrt[3]2)(x^2-\sqrt[3]2x+\sqrt[3]4).$$

The answer would be $$\frac{1}{-x^4-2x}=-\frac{1}{2x}+\frac{2x-\sqrt[3]2}{6((x^2-\sqrt[3]2x+\sqrt[3]4)}+\frac{1}{6(x+\sqrt[3]2)}.$$

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