- #1

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^{3}+ 1)(x

^{3}- 1) or any other problem where the degree of the denominator is at least 2 larger than the numerator. how do I do this?

- Thread starter LusTRouZ
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- #1

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- #2

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No, it doesn't.

>

Note that 1/(X^7 + x) = 1/x * 1/(1+x^6)

= 1/x * (1 -x^6 + x^12 - x^18 + ...)

=1/x - x^5 + x^11 - x^17 + ...

Then the integral is

C + log(x) - x^6/6 + x^12/12 - x^18/18 + ...

the series part is clearly a logarithm expansion, so it's

C + logx - log(1+x^6)/6

1/(X^7 + x) = x^-7 * 1/(1+x^-6)

= x^-7 * (1 - x^-6 + x^-12 + x^-18 + ...)

=x^-7 - x^-13 + x^-19 - ...

Then the integral is

C -x^-6/6 + x^-12/12 - x^-18/18 + ...

i.e. it's

C - log(1+x^-6)/6

which is equal to the above solution.

1/(X^7 + x) = x^-7/(1+x^-6)

and the solution

is C - log (1+x^-6)/6 based on inspection (it clearly produces the correct derivative).

- #3

Mark44

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Your factorization is incorrect. x^{3}+ 1)(x^{3}- 1) or any other problem where the degree of the denominator is at least 2 larger than the numerator. how do I do this?

[tex]\frac{1}{x^7 + x} = \frac{1}{x(x^6 + 1)} = \frac{1}{x(x^2 + 1)(x^4 - x^2 + 1)}[/tex]

You can break up that quartic into x

- #4

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is there any way of doing this with partial fractions?

- #5

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1/(x+x^7) = 1/x * 1/(1+x^6) = 1/x * 1/((1+ax)*(1+bx)*(1+cx)*(1+dx)*(1+ex)*(1+fx))

where a,b,c,d,e,f are the six roots of unity exp(2 i pi N/6) for N = 0...5

= 1/(6x) * (1/(1+ax) + 1/(1+bx) + ... + 1/(1+fx))

This integrates to

C - (1/6)[ log(a+1/x) + log(b + 1/x) + log(c+1/x) + ... + log(f+1/x)]

which is the same as three other solutions listed above.

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