Question

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

Answer 1

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

Answer 2

\[x ^{2}/(1-x)(1+x)=A/(1-x)+B(1+x)\]

Answer 3

its B/(1+x)

Answer 4

ya?

Answer 5

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.

Answer 6

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

Answer 7

so, one thing u can do is, resolve 1/(1-x^2) into partial fractions and then multiply x^2 to both terms.

Answer 8

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

Answer 9

or u can do that ^

Answer 10

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

Answer 11

Hmmm... -1 -1 = -2?!
Shouldn't it be
\[\frac{-(-x^2)}{(1-x^2) } = \frac{-(1-x^2)+1}{1-x^2} = -1+\frac{1}{1-x^2}\]

Answer 12

yes, sorry.

Answer 13

but u got the point ?

Answer 14

@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

Answer 15

@hartnn Yes! When can I be as smart as you???!!!!!

Answer 16

with just a little bit of practice :P

Answer 17

\[\frac{x^2}{1-x^2}=x^2\times\frac{1}{1-x^2} = x^2\times\frac{1}{(1-x)(1+x)}\]

\[\frac{1}{(1-x)(1+x)}=\frac A{(1−x)}+\frac B{(1+x)}=\frac{A(1+x)+B(1-x)}{(1-x)(1+x)}\]

\[1=A(1+x)+B(1-x)\]

for \(x=1\)

\(1=2A\)
\(A=1/2\)

for \(x=-1\)

\(1=-2B\)
\(B=-1/2\)

\[\frac{1}{(1-x)(1+x)}=\frac{1/2}{(1−x)}-\frac {1/2}{(1+x)}\]
\[\frac{x^2}{(1-x)(1+x)}=\frac{x^2}{2(1−x)}-\frac {x^2}{2(1+x)}\]

Answer 18

Excuse.. me....

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

Answer 19

your right

Answer 20

http://www.wolframalpha.com/input/?i=Resolve+into+partial+fraction%3A+x%5E2+%2F+%281-x%5E2%29
I don't know how to get that -1/(2(1-x))

Answer 21

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

Answer 22

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

Answer 23

im not sure about the -1

Answer 24

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,
\[\frac{x^2}{1-x^2}=-1+\frac{1}{2(1+x)}+\frac{1}{2(1-x)}\]

Answer 25

A = B = 1/2

great work

@RolyPoly

Answer 26

1-x= -(x-1)
again

Answer 27

what u have done is correct.

Answer 28

:'(
Last but not least...
\[\frac{x^2}{1-x^2}=-1+\frac{1}{2(1+x)}-\frac{1}{2(x-1)}\]

Answer 29

It needs tons of thousands of millions of practice!! :'(
Thanks again!!!

Answer 30

welcome ^_^