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# C4. :) watch

1. Hi,

How do I set up this partial fraction?

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

Is it going to be three fractions? I'm just stuck on what goes on the bottom of each of them.

I am then asked to integrate the above. For this would I integrate the partial fractions?

Thanks.
2. (Original post by apo1324)
Hi,

How do I set up this partial fraction?

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

Is it going to be three fractions? I'm just stuck on what goes on the bottom of each of them.

I am then asked to integrate the above. For this would I integrate the partial fractions?

Thanks.
On the bottom you'd have x, x^2 and (x-1)
then use algabreic symbols on top and solve for the above
3. (Original post by apo1324)
Hi,

How do I set up this partial fraction?

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

Is it going to be three fractions? I'm just stuck on what goes on the bottom of each of them.

I am then asked to integrate the above. For this would I integrate the partial fractions?

Thanks.
as the greatest power on the bottom is 3 then I believe it splits into three fractions.
Should be x-1, x^2 and x^3...I think.

Apologies if that turns out bad.

Yeah forget this, the above post was right :P
4. (Original post by apo1324)
Hi,

How do I set up this partial fraction?

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

Is it going to be three fractions? I'm just stuck on what goes on the bottom of each of them.

I am then asked to integrate the above. For this would I integrate the partial fractions?

Thanks.
Ok well because the denominator will expand out to an x^3 and the numerator's highest power is only x^2 its not an improper fraction

So it will split in to 3

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

which will be equal to 3x^2 + 2x + 1 then solve for A B C then integrate
5. Yeah and you integrate the partial fractions
6. Thank you all.

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