# IntegrationWatch

#1
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
0
4 years ago
#2
(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?
0
4 years ago
#3
(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
then write 4x+1 as a multiple of 2x-2 (the derivative of the denominator)+ a constant
Can you proceed from there?
0
4 years ago
#4
(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?
0
#5
(Original post by brianeverit)
First step is writing
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?
0
4 years ago
#6
(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.
0
4 years ago
#7
(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

0
4 years ago
#8
(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.
0
#9
(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!
0
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