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Resolve into partial fraction \[\frac{x^2}{1-x^2}\]

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\[\frac{x^2}{1-x^2} = \frac{x^2}{(1-x)(1+x)}\]How to continue??
\[x ^{2}/(1-x)(1+x)=A/(1-x)+B(1+x)\]
its B/(1+x)

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But that doesn't seems to work.. Here's what I mean: A(1+x) + B(1-x) = x^2 (A-B)x +(A+B) = x^2 => A-B =0 , A+B =0 Not reasonable.
The degree of numerator must strictly be less than degree of denominator.
so, one thing u can do is, resolve 1/(1-x^2) into partial fractions and then multiply x^2 to both terms.
\(\large \frac{-(-x^2)}{(1-x^2) } = \frac{-(1-x^2)-1}{1-x^2} = -1-\frac{1}{1-x^2}\)
or u can do that ^
A(1+x) + B(1-x) = x^2 for x= -1 2 B = x^2 B = x^2/2 for x= 0 A + B = x^2 for x= 1 2A = x^2 A = x^2/2
Hmmm... -1 -1 = -2?! Shouldn't it be \[\frac{-(-x^2)}{(1-x^2) } = \frac{-(1-x^2)+1}{1-x^2} = -1+\frac{1}{1-x^2}\]
yes, sorry.
but u got the point ?
@UnkleRhaukus A(1+x) + B(1-x) = x^2 for x= -1 2 B = (-1)^2 B = 1/2 for x= 0 A + B = 0 for x= 1 2A = 1^2 A = 1/2 And it doesn't work here :S
@hartnn Yes! When can I be as smart as you???!!!!!
with just a little bit of practice :P
\[\frac{x^2}{1-x^2}=x^2\times\frac{1}{1-x^2} = x^2\times\frac{1}{(1-x)(1+x)}\] \[\frac{1}{(1-x)(1+x)}=\frac A{(1−x)}+\frac B{(1+x)}=\frac{A(1+x)+B(1-x)}{(1-x)(1+x)}\] \[1=A(1+x)+B(1-x)\] for \(x=1\) \(1=2A\) \(A=1/2\) for \(x=-1\) \(1=-2B\) \(B=-1/2\) \[\frac{1}{(1-x)(1+x)}=\frac{1/2}{(1−x)}-\frac {1/2}{(1+x)}\] \[\frac{x^2}{(1-x)(1+x)}=\frac{x^2}{2(1−x)}-\frac {x^2}{2(1+x)}\]
Excuse.. me.... When x =-1 1 = A(1+(-1)) + B(1-(-1)) 1 = 0 + 2B B = 1/2?
your right I don't know how to get that -1/(2(1-x))
Such simple maths is killing me :\ Thanks!!!!!!!!!
im not sure about the -1
Here it goes~ \[\frac{x^2}{1-x^2}\]\[=\frac{-(-x^2)}{1-x^2}\]\[=\frac{-(1-x^2)+1}{1-x^2}\]\[=-1+\frac{1}{1-x^2}\] \[\frac{1}{1-x^2}=\frac{A}{1+x}+\frac{B}{1-x}\]A(1-x)+B(1+x)=1 A = B = 1/2 So, \[\frac{1}{1-x^2}=\frac{1}{2(1+x)}+\frac{1}{2(1-x)}\] Thus, \[\frac{x^2}{1-x^2}=-1+\frac{1}{2(1+x)}+\frac{1}{2(1-x)}\]
A = B = 1/2 great work @RolyPoly
1-x= -(x-1) again
what u have done is correct.
:'( Last but not least... \[\frac{x^2}{1-x^2}=-1+\frac{1}{2(1+x)}-\frac{1}{2(x-1)}\]
It needs tons of thousands of millions of practice!! :'( Thanks again!!!
welcome ^_^

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