Write Expression In Partial Fraction

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November 8th 2011, 06:42 PM
gilagila

Write Expression In Partial Fraction

x^2/(x-1)(x+1) = x^2/x^2-1 = 1,-1

Therefore,
x^2/(x-1)(x+1) = 1 + -1/(x-1)(x+1)

-1/(x-1)(x+1) = A/(x-1) + B/(x+1)
-1 = Ax + A + Bx - B
-1 = (A+B)x +A - B
A+B = 0, B = -A
-1 = A - B = A - (-A) = 2A
-1 = 2A, A = -1/2,
B = -(-1/2) = 1/2

Therefore,
x^2/(x-1)(x+1) = 1 - 1/2(x-1) + 1/2(x+1)

But somehow the answer given to me is
x^2/(x-1)(x+1) = 1 + 1/2(x-1) - 1/2(x+1)

Not sure which step I doing wrong, please help me on this. Thank you.
November 8th 2011, 08:20 PM
Soroban

Re: Write Expression In Partial Fraction

Hello, gilagila!

Quote:

$\frac{x^2}{(x-1)(x+1)}$

We have: . $\frac{x^2}{x^2-1} \;=\;\frac{x^2 - 1 + 1}{x^2-1} \;=\;\frac{x^2-1}{x^2-1} + \frac{1}{x^2-1} \;=\;1 + \frac{1}{x^2-1}$

Then: . $\frac{1}{(x-1)(x+1)} \;=\;\frac{A}{x-1} + \frac{B}{x+1}$

. . . . . . $1 \;=\;A(x+1) + B(x-1)$

Let $x = 1\!:\;\;1 \;=\;A(2) + B(0) \quad\Rightarrow\quad A \,=\,\tfrac{1}{2}$

Let $x = \text{-}1\!:\;\;1 \;=\;A(0) + B(\text{-}2) \quad\Rightarrow\quad B \,=\,\text{-}\tfrac{1}{2}$

Therefore: . $\frac{x^2}{(x-1)(x+1)} \;=\; 1 + \frac{\frac{1}{2}}{x-1} - \frac{\frac{1}{2}}{x+1}$
November 9th 2011, 05:41 PM
gilagila

Re: Write Expression In Partial Fraction

Quote:

Originally Posted by Soroban

$\frac{x^2}{x^2-1} \;=\;\frac{x^2 - 1 + 1}{x^2-1} \;$

Just want to check, how you change to this ??
November 9th 2011, 09:35 PM
sbhatnagar

Re: Write Expression In Partial Fraction

Quote:

Originally Posted by gilagila

Just want to check, how you change to this ??

$\frac{x^2}{x^2-1}=\frac{x^2-1+1}{x^2-1}$

He subtracted 1 from the numerator and at the same time added it. This does not make any "change" to expression.

http://latex.codecogs.com/gif.latex?x^2 $=$ http://latex.codecogs.com/gif.latex?x^2 http://latex.codecogs.com/gif.latex?...pace;-1+1
November 9th 2011, 10:31 PM
gilagila

Re: Write Expression In Partial Fraction

oic, thanks for explanation