baldymcgee6
How to setup this integral with partial fractions..?
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baldymcgee6
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|dw:1358744328225:dw|
baldymcgee6
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x^7
baldymcgee6
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just have to set it up, don't have to actually integrate it. :)
hartnn
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x(x^6-1)
=x(x^3+1)(x^3-1)
baldymcgee6
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\[\LARGE{\int\limits \frac{ 1 }{ x^7 - x }dx}\]
kirbykirby
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|dw:1358744642346:dw|
hartnn
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do you know general formulas for a^3+b^3 and a^3-b^3..
baldymcgee6
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yes
baldymcgee6
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so how would this spit into partials?
hartnn
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if there is quadratic equation in denominator, which cannot be factorized, then in numerator, it'll be linear expression, Ax+B...
kirbykirby
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For x^3-1 and x^3+1 you can use Long polynomial division by finding a factor (x-a) where a is a root of the polynomial
hartnn
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so, whats x^3+1=... ?
x^3-1=.... ?
kirbykirby
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I never learnt the general formula for a^3+b^3. I think it's easier to do polynomial division because it will work for higher powers
baldymcgee6
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x^3 +1 = (x+1)(x^2-x+1)
hartnn
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so, above x^2-x+1 will come a linear expression like Ax+B
and above linear terms like x, x(x+1), constants like C,D,E ... will come.
baldymcgee6
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@kirbykirby polynomial division of what and what?
baldymcgee6
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x^3 -1 = (x-1)(x^2+x+1)
baldymcgee6
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okay, i'll try this
hartnn
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so, you have 5 factors, out of which 2 is quadratic.
kirbykirby
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for x^3+1, you find the factors of the constant (here, its 1) and find which one = 0 when you plug it into your x-terms. So, -1 works: ((-1)^3+1) = 0. So, the factor that divides the polynomial is always of the form (x-a). So, (x-(-1)) = (x+1).
Then divide x^3+1 into x+1
baldymcgee6
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|dw:1358745155474:dw|
hartnn
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\(\checkmark \)
baldymcgee6
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@kirbykirby that is useful to know rather than memorizing these formulas.
baldymcgee6
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thanks for the help guys!
kirbykirby
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No problem :)