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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)

ya?

The degree of numerator must strictly be less than degree of denominator.

\(\large \frac{-(-x^2)}{(1-x^2) } = \frac{-(1-x^2)-1}{1-x^2} = -1-\frac{1}{1-x^2}\)

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

yes, sorry.

but u got the point ?

with just a little bit of practice :P

Excuse.. me....
When x =-1
1 = A(1+(-1)) + B(1-(-1))
1 = 0 + 2B
B = 1/2?

your right

(x-1)=-(1-x)

Such simple maths is killing me :\
Thanks!!!!!!!!!

im not sure about the -1

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 ^_^