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010c
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Question: integrate (5x^2-7x+26)/(x^2-2x+5)

Can anybody help me on this please? I know the answer involves tan^-1(x-1/2) but I don't know what else to do
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Smaug123
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(Original post by 010c)
Question: integrate (5x^2-7x+26)/(x^2-2x+5)

Can anybody help me on this please? I know the answer involves tan^-1(x-1/2) but I don't know what else to do
If you're familiar with contour integration, it falls immediately to that. Otherwise: you have a top-heavy fraction. What can we do with top-heavy fractions?
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brianeverit
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(Original post by 010c)
Question: integrate (5x^2-7x+26)/(x^2-2x+5)

Can anybody help me on this please? I know the answer involves tan^-1(x-1/2) but I don't know what else to do
First step is writing 5x^2-7x+26\text{ as }5(x^2-2x+5)+3x+1
then write 4x+1 as a multiple of 2x-2 (the derivative of the denominator)+ a constant
Can you proceed from there?
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davros
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(Original post by Smaug123)
If you're familiar with contour integration, it falls immediately to that.
I haven't woken up yet this morning, but how is contour integration going to help with an indefinite integral?
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010c
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(Original post by brianeverit)
First step is writing 5x^2-7x+26\text{ as }5(x^2-2x+5)+3x+1
then write 4x+1 as a multiple of 2x-2 (the derivative of the denominator)+ a constant
Can you proceed from there?

Where has 4x + 1 come from?

I now have the integral of 5 + the integral of (3x +1)/(x^2-2x+5) but 3x+1 isn't a multiple of the 2x-2?
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davros
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(Original post by 010c)
Where has 4x + 1 come from?

I now have the integral of 5 + the integral of (3x +1)/(x^2-2x+5) but 3x+1 isn't a multiple of the 2x-2?
You'll want to repeat the trick you did earlier by writing 3x+1 = 3/2(2x - 2) + some constant leaving you with 2 more integrals to do: one will give a logarithm and the other will be a standard inverse tangent.
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TenOfThem
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(Original post by 010c)
Where has 4x + 1 come from?

I now have the integral of 5 + the integral of (3x +1)/(x^2-2x+5) but 3x+1 isn't a multiple of the 2x-2?
You need to keep going to get

 5 + \dfrac{1}{2} \times \dfrac{6x-6}{x^2-2x+5} +\dfrac{4}{x^2-2x+5}
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Smaug123
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(Original post by davros)
I haven't woken up yet this morning, but how is contour integration going to help with an indefinite integral?
Oops. Memo to self: don't answer questions immediately after waking up or just before going to bed.
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010c
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(Original post by davros)
You'll want to repeat the trick you did earlier by writing 3x+1 = 3/2(2x - 2) + some constant leaving you with 2 more integrals to do: one will give a logarithm and the other will be a standard inverse tangent.

Ah I see now, thank you for your help!
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