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.
On
I'm guessing that this problem is homework 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
$$1=Ax^3+A\sqrt{2}x^2+Ax+Bx^2+B\sqrt{2}x+B+Cx^3-C\sqrt{2}x^2+Cx+Dx^2-D\sqrt{2}x+D$$
My favorate partial fraction decomposition.
$$\frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1} -\frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1} \right)$$