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Then just integrate it term by term to get the indefinite integral in an infinite series form.
Starting with the 1/x^2 term the coefficients in the series repeat in blocks of 10 in the pattern
1,-1,1,0,0,-1,1,-1,0,0. That is what the 1/(x^10n) handles. So the next 6 terms in the series
would have exponents on the bottom of 12,13,14,17,18,19 and the next 6 would have
22,23,24,27,28,29 for the exponents on the bottom.
The signs in each block of six terms would continue to be 1, -1, 1, -1, 1, -1.
Has anyone got a closed form for the integral? I'd love to see it. If x^5+1 is divided by x-1 I get
x(x+1)(x-1)(x^2+1) + (x+1)/(x-1) but I haven't been able to use this to get a closed form.
Good luck with it!
The integral can be solved through partial fractions, but it is a nasty factorisation!
Sorry, you are right. Selective blindness again.
It is a factor of the denominator only.
It is not.
(x + 1) is a factor of top and bottom.