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You are not logged in. #3 20121129 20:15:16
Re: Indefinite integrationHi jacks; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #4 20121130 00:46:46
Re: Indefinite integration
It is not. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #5 20121130 02:19:20
Re: Indefinite integrationIt is a factor of the denominator only. 'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.' 'God exists because Mathematics is consistent, and the devil exists because we cannot prove it' 'Who are you to judge everything?' Alokananda #7 20121130 05:29:25
Re: Indefinite integrationThe integral can be solved through partial fractions, but it is a nasty factorisation! The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #8 20121130 08:20:54
Re: Indefinite integrationHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #9 20121130 14:38:17
Re: Indefinite integrationHi! 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 x1 I get x(x+1)(x1)(x^2+1) + (x+1)/(x1) but I haven't been able to use this to get a closed form. Good luck with it! Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional). LaTex is like painting on many strips of paper and then stacking them to see what picture they make. 