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# Partial Fractions watch

1. away when we did this. I think I understand the concept.

3x-2/2x+1 is a partial fraction
3x-2/(2x-1)^3 is repeated.

I have a question. x^2+1/x(2x^2+1) : Express in partial fractions.
How do you go about doing this. Is it quadratic? If so how do you work it out.

I've got someting saying put it in ax+b/something + c/something

so do you do:

ax+b/2x^2+1 + c/x

I then worked out C to be 1 but how do you get A and B. Am I on the right lines at all ????

Any help would be appreciated.

Thanks
2. (x^2 + 1) / [x (2x^2 + 1)]
= (ax + b) / (2x^2 + 1) + c/x.

Multiply through by x (2x^2 + 1) to get

(x^2 + 1) = (ax + b)x + c(2x^2 + 1).

Compare constants: 1 = c.
Compare coefficients of x: 0 = b.
Compare coefficients of x^2: 1 = a + 2c, so a = -1.

So

(x^2 + 1) / [x (2x^2 + 1)]
= -x / (2x^2 + 1) + 1/x.
3. Thanks

One more question.

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

My final answer came out as

3/5(x+2) - 3/5(2x+1) - 4/25(x^2+1)

The first 2 parts were right but the last bit was wrong with the real answer being

3/5(x+2) - 3/5(2x+1) + (x-1)/5(x^2+1)

I initially put it in the form

Ax+b/x^2+1 + c/x+2 + d/2x+1

I think this is right but I worked out B to be 0 and is this where I went wrong?

Thanks
4. (Original post by Alex H)
x^3-1/(x+2)(2x+1)(x^2+1)
...
I initially put it in the form

Ax+b/x^2+1 + c/x+2 + d/2x+1

I think this is right but I worked out B to be 0 and is this where I went wrong?
No, ur method were right, but maybe u had some mistakes on calculation.
Using ur way, I have:
(ax+b)(x+2)(2x+1) + c(x^2+1)(2x+1) + d(x^2+1)(x+2) = x^3 - 1
let x = -2, we will have c = 3/5
let x = -1/2, we have d = -3/5
let x = 0, we have:
2b + 3/5 - 2*(3/5) = -1
2b = -1 + 3/5 = -2/5
b = -1/5
let x = 1 we have:
9(a+b) + 6*(3/5) - 6*(3/5) = 0
a+b = 0
a = -b = 1/5
3/5(x+2) - 3/5(2x+1) + (x-1)/5(x^2 +1)
Have fun.
5. I think it was working out the value for B that I mssed up on. Thanks!!

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