Mathematics: Partial Fractions

Mansize_tissue

Mansize_tissue

Mansize_tissue

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'Any maths guys in here who want to give me a hand?

Express the following as a sum of partial fractions:

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


Here's what I've done:

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

So...

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

From there, by substituting x = -0.5, I got A = 6
Then I equated the coefficients and got B = -3 and C = 2

However, apparently this is incorrect and I can't work out where I've gone wrong. Any help?
 
Mansize_tissue said:
'Any maths guys in here who want to give me a hand?

Express the following as a sum of partial fractions:

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


Here's what I've done:

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

So...

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

From there, by substituting x = -0.5, I got A = 6
Then I equated the coefficients and got B = -3 and C = 2

However, apparently this is incorrect and I can't work out where I've gone wrong. Any help?
Click to expand...

Cant recall all of my C4 but i think you have used the wrong partial fraction for this type of equation.

Equations with a denominator of (x^2 + q)(px + r) should equal ((Ax+B)/(x^2 + q)) + (C/(px + r))

So Ax + B and C

rather than A and B and C


Solution:

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

=(Ax + B)/(x^2) + (C)/(2x+1)

Multiply by denominators to give;

x^2 - 1 = (Ax+B)(2x+1) + (C)(x^2)

When x = 0; -1 = Ax+B (1)
When x = 0.5; -0.75 = 0.25C

C=-3

Equating quadratic coefficients;

1 = 2A + C
A = 2

Sub A=2 into (1)

gives B = -3
 
Last edited:
That's what I thought, >:|sh4d0w|:<.

However, this is the answer in the book:

2/x - 1/x^2 - 3/(2x + 1)

So there are 3 separate fractions, apparently. This doesn't make any sense. :confused:

Thank you for the prompt reply, by the way.
 
Oh, I see now. Both methods work I think, I was just doing it the very long way. The reason my method didn't work I think was because I needlessly multiplied each term by x when I didn't have to in order to make the denominators the same.

Thank you very much for your help.

However, I'm going to have to be picky:
>:|sh4d0w|:< said:
When x = 0; -1 = Ax+B (1)
Click to expand...
If x = 0 won't Ax = 0?
 
Last edited:
x^2-1=A(2x+1)+B(x^2)+C(x(2x+1)

x=0.5 -> 0.25-1=0A+0.25B+0C -> B=4(-0.75) -> B=-3

x=0 -> -1=1A+0B+0C -> A=-1

equating coefficients of x^2 -> -1=B+2C but B=-3, so 2C=4 -> C=2

then (x^2-1)/(x^2)(2x+1)=2/x-1/x^2-3/(2x+1)

EDIT: sorry bit slow, shame ocuk forums don't have equation editor.
 
Last edited:
Their are 3 separate fractions because their are 3 solutions for x in a cubic function, which (x^2)(2x+1) is. (Can be expressed as 2x^3+x^2).
 
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