Is that correct? Resolve into partial fraction\[\frac{x^2}{x^4-1}\]\[\frac{x^2}{x^4-1}=\frac{x^2}{(x^2+1)(x-1)(x+1)}\] \[A(x-1)(x+1) + B(x^2+1)(x-1) + C(x^2+1)(x+1) = x^2\]\[(B+C) x^3+(A-B+C)x^2 +(B+C)x -A-B+C =x^2\]So, B+C = 0, A-B+C=1, A+B-C = 0 And I got A = 1/2, B=-1/4, C=1/4 So, \[\frac{x^2}{x^4-1} = \frac{1}{2(x^2+1)} - \frac{1}{4(x+1)} +\frac{1}{4(x-1)}\]

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Is that correct? Resolve into partial fraction\[\frac{x^2}{x^4-1}\]\[\frac{x^2}{x^4-1}=\frac{x^2}{(x^2+1)(x-1)(x+1)}\] \[A(x-1)(x+1) + B(x^2+1)(x-1) + C(x^2+1)(x+1) = x^2\]\[(B+C) x^3+(A-B+C)x^2 +(B+C)x -A-B+C =x^2\]So, B+C = 0, A-B+C=1, A+B-C = 0 And I got A = 1/2, B=-1/4, C=1/4 So, \[\frac{x^2}{x^4-1} = \frac{1}{2(x^2+1)} - \frac{1}{4(x+1)} +\frac{1}{4(x-1)}\]

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|dw:1350710674124:dw|
@RolyPoly good to go.
@zepdrix Yes! That should be the case, but I can't get the answer by solving that equation :(

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Hmmm :c
Continue from that step; (A+C+D)x^3 + (B+C-D)x^2 + (-A+C+D) + (-B+C-D) = x^2 So, A+C+D = 0 --- (1) (B+C-D) = 1 --- (2) -A+C+D =0 --- (3) -B+C-D = 0 --- (4) (1)-(3): A=0 (3)-(4): -A+B = 0 => B = 0 Here comes the problem!
|dw:1350711031547:dw| Hmmmm we have 4 equations and 4 unknowns. This should be solvable. It might be a bit of work though :o
Oh hmm
|dw:1350711417164:dw| Hmm maybe just try combining them a different way? <:O I came up with this, let me know if my math was incorrect.
I'm also getting A=0, I don't think that's a problem though. Since I was able to get a B term.
Oh i see the problem, (3)-(4) = (-A+C+D) - (-B+C-D) = -A + B +2D = 0
Uh-huh! My mistake!! Thanks :)
Yah these things suck :) lol
(2) - (4): B = 1/2 Yay!

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