anonymous
  • anonymous
pls sove this by partial fraction method(x^2+1)/x(x-1)(x+1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • 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
  • anonymous
Here A= -1, B=2, C=0
anonymous
  • 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
  • anonymous
i got a=1 b=1 and c=1
anonymous
  • anonymous
You may be right but I found A by x=0 and B & C by considering x = 1 and x = -1.
anonymous
  • anonymous
and my text says 1/(x-1)+1/(x+1)-1/x
anonymous
  • anonymous
We are both correct.
anonymous
  • anonymous
o.k will try it once again ....thanks alot
anonymous
  • anonymous
i think A=-1,,B=C=1
anonymous
  • anonymous
so the text book says A=-1 ok
anonymous
  • anonymous
i mean A=1... ok
anonymous
  • anonymous
no A= -1
anonymous
  • anonymous
B=C=1
anonymous
  • anonymous
yeah A=-1
anonymous
  • anonymous
ok as long as you got the correct solution and answer....LOL
anonymous
  • 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
  • anonymous
mhhhmm thanks both of you
anonymous
  • anonymous
but 0=Bx - Cx therefore 0=x(B-C) B=C ..If B=1=C
anonymous
  • anonymous
Where did you get 0=Bx-Cx from?
anonymous
  • anonymous
expand x^2 + 1 = A(x-1)(x+1) + Bx(X+1) +Cx(x-1)
anonymous
  • anonymous
That is incorrect: C is the constant term. There is no Cx.
anonymous
  • anonymous
=Ax^2-A + Bx^2+Bx+Cx^2-Cx
anonymous
  • anonymous
mark the quation you have given might be correct ....can please elaborate for the value of A
anonymous
  • anonymous
WHAT VALUE OF "X" DID YOU TAKE TO FIND THE VALUE OF "A"
anonymous
  • anonymous
No; (x+1)/(x^2-1) - 1/x does not work. But (2x)/(x^2-1) - 1/x does.
anonymous
  • anonymous
from (x^2 +1) A B C ------- = --- + --- + ---- x(x-1)(x+1) x x-1 x+1
anonymous
  • anonymous
I was not using that anzats lol. rahulyadav was correct.
anonymous
  • anonymous
x^2 + 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)
anonymous
  • anonymous
Here A=-1 and B=C=1.
anonymous
  • anonymous
mark o. to find value of A what value of x did you take???
anonymous
  • anonymous
Try x=0.
anonymous
  • anonymous
When x=0, 1=A(-1)(1) <=> A=-1.
anonymous
  • 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
  • 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
  • anonymous
That does not matter if you have already found A using x=0.
anonymous
  • anonymous
0=Bx-Cx 0=x(B-C) 0=(B-C)... B=C
anonymous
  • anonymous
so finally both the values work for me but which value should i use?...x=1,x=-1,or x=0????
anonymous
  • anonymous
All three.
anonymous
  • anonymous
yes it works also
anonymous
  • 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
  • anonymous
thanks both of you for giving so much time to this problem
anonymous
  • anonymous
i just did it a long way.... and you can do it a short cut method like what shaun did....
anonymous
  • anonymous
ok welcome and good luck
anonymous
  • 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
  • anonymous
yes correct you can use 3 lines in equal sign

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