pls sove this by partial fraction method(x^2+1)/x(x-1)(x+1)

- anonymous

pls sove this by partial fraction method(x^2+1)/x(x-1)(x+1)

- jamiebookeater

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- anonymous

Start by expanding (x-1)(x+1) to get x^2-1. Then try this anzats (or estimation): A/x + (Bx+c)/(x^2-1).
You can then cross multiply to get: \[x ^{2}+1=A(x^{2}-1)+x(Bx+C)\]
Is that enough to work with? I don't know what you mean by "solve" in this context but if you try different x values in the above - or expand out the LHS - you can find A, B & C, which you just sub back in the anzats to find the partial fractions.
I hope that helps :)

- anonymous

Here A= -1, B=2, C=0

- anonymous

i have tried and got the\[(1\div x)+1\div(x+1)+1\div(x-1)\] answer ...is this right

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- anonymous

i got a=1 b=1 and c=1

- anonymous

You may be right but I found A by x=0 and B & C by considering x = 1 and x = -1.

- anonymous

and my text says 1/(x-1)+1/(x+1)-1/x

- anonymous

We are both correct.

- anonymous

o.k will try it once again ....thanks alot

- anonymous

i think A=-1,,B=C=1

- anonymous

so the text book says A=-1 ok

- anonymous

i mean A=1... ok

- anonymous

no A= -1

- anonymous

B=C=1

- anonymous

yeah A=-1

- anonymous

ok as long as you got the correct solution and answer....LOL

- anonymous

When x=0, 1=A(-1) <=> A=-1.
When x=1, 2=A(1-1) +(B+C) and when x=-1, 2=A(1-1) +(B-C) <=> C=0 & B=2.

- anonymous

mhhhmm thanks both of you

- anonymous

but 0=Bx - Cx therefore
0=x(B-C)
B=C ..If B=1=C

- anonymous

Where did you get 0=Bx-Cx from?

- anonymous

expand
x^2 + 1 = A(x-1)(x+1) + Bx(X+1) +Cx(x-1)

- anonymous

That is incorrect: C is the constant term. There is no Cx.

- anonymous

=Ax^2-A + Bx^2+Bx+Cx^2-Cx

- anonymous

mark the quation you have given might be correct ....can please elaborate for the value of A

- anonymous

WHAT VALUE OF "X" DID YOU TAKE TO FIND THE VALUE OF "A"

- anonymous

No; (x+1)/(x^2-1) - 1/x does not work. But (2x)/(x^2-1) - 1/x does.

- anonymous

from (x^2 +1) A B C
------- = --- + --- + ----
x(x-1)(x+1) x x-1 x+1

- anonymous

I was not using that anzats lol. rahulyadav was correct.

- anonymous

x^2 + 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)

- anonymous

Here A=-1 and B=C=1.

- anonymous

mark o.
to find value of A what value of x did you take???

- anonymous

Try x=0.

- anonymous

When x=0, 1=A(-1)(1) <=> A=-1.

- anonymous

i am not confident about the value of which i got cause the coefficient of A ALWAYS COMES DOWN TO "0" IF I PUT X=1 OR X=-1

- anonymous

=A(x^2 -1)+B(x^2 +x) +C(x^2-x)
x^2+1 = Ax^2 - A + Bx^2 + Bx + Cx^2 -Cx
1=-A......A=-1

- anonymous

That does not matter if you have already found A using x=0.

- anonymous

0=Bx-Cx
0=x(B-C)
0=(B-C)...
B=C

- anonymous

so finally both the values work for me but which value should i use?...x=1,x=-1,or x=0????

- anonymous

All three.

- anonymous

yes it works also

- anonymous

The purpose of using x values is to discover what A, B & C are. The anzats mark put up is called an identity. It works for all x values.

- anonymous

thanks both of you for giving so much time to this problem

- anonymous

i just did it a long way.... and you can do it a short cut method like what shaun did....

- anonymous

ok welcome and good luck

- anonymous

Strictly speaking, the "=" sign should have three lines to make that clear. When writing an anzats for partial fractions, try using the three lines to remind you that it holds for any x. And you're welcome :)

- anonymous

yes correct you can use 3 lines in equal sign

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