Question

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

Answer 1

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 :)

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

We are both correct.

Answer 8

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

Answer 9

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

Answer 10

so the text book says A=-1 ok

Answer 11

i mean A=1... ok

Answer 12

no A= -1

Answer 13

B=C=1

Answer 14

yeah A=-1

Answer 15

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

Answer 16

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.

Answer 17

mhhhmm thanks both of you

Answer 18

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

Answer 19

Where did you get 0=Bx-Cx from?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

Try x=0.

Answer 32

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

Answer 33

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

Answer 34

=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

Answer 35

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

Answer 36

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

B=C

Answer 37

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

Answer 38

All three.

Answer 39

yes it works also

Answer 40

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.

Answer 41

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

Answer 42

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

Answer 43

ok welcome and good luck

Answer 44

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 :)

Answer 45

yes correct you can use 3 lines in equal sign