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I need some help figuring out how to decompose $\displaystyle\frac{1}{x^4+1}$ into partial fractions.

This is what I have done so far: $$\frac{1}{x^4+1} = \frac{1}{(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)}$$

From there, I do not know how to decompose it into partial fractions, or if I even set it up correctly.

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    $\begingroup$ So far so good. I think the next step is to write $$\frac1{x^4+1} = \frac{Ax+B}{x^2-\sqrt2x+1} + \frac{Cx+D}{x^2+\sqrt2x+1}$$ and then try to solve for $A,B,C,D$. $\endgroup$ – MJD Jun 10 '14 at 0:30
  • $\begingroup$ I tried that, and I could not get valid equations. Assuming I computed everything correctly, I got the following equations: $B+D=0$ and $B+D=1$, which is invalid. $\endgroup$ – A is for Ambition Jun 10 '14 at 0:32
  • $\begingroup$ That means that you made a mistake in your arithmetic. Check the arithmetic carefully, and if you can't find the error, reproduce your steps here. $\endgroup$ – MJD Jun 10 '14 at 0:33
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My favorate partial fraction decomposition.

$$\frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1} -\frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1} \right)$$

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I'm guessing that this problem is so I'll provide a step forward from where you're stuck to see if it'll help you.

Currently, you have this factored into two distinct irreducible quadratics, each of which gets a term in the form $\frac{Ax+B}{ax^2+bx+c}$. Therefore, you should set up the equation to be:

$$\frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}=\frac{Ax+B}{x^2-\sqrt{2}x+1}+\frac{Cx+D}{x^2+\sqrt{2}x+1}$$

Continue by determining the values of A, B, C, and D you need.

I tried that, and I could not get valid equations. Assuming I computed everything correctly, I got the following equations: B+D=0 and B+D=1, which is invalid.

Those aren't the equations you're looking for. If you multiply both sides by $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$, you will get:

$$1=(Ax+B)(x^2+\sqrt{2}x+1)+(Cx+D)(x^2-\sqrt{2}x+1)$$ Multiply out to get

$$1=Ax^3+A\sqrt{2}x^2+Ax+Bx^2+B\sqrt{2}x+B+Cx^3-C\sqrt{2}x^2+Cx+Dx^2-D\sqrt{2}x+D$$

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  • $\begingroup$ Right, sorry. My arithmetic was incorrect. $\endgroup$ – A is for Ambition Jun 10 '14 at 0:40

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