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jacks
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## #2 2012-11-29 20:00:08

bob bundy
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### Re: Indefinite integration

(x + 1) is a factor of top and bottom.

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #3 2012-11-29 20:15:16

bobbym

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### Re: Indefinite integration

Hi jacks;

The answer for that is huge, are you supposed to do that by hand?

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Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #4 2012-11-30 00:46:46

anonimnystefy
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### Re: Indefinite integration

#### bob bundy wrote:

(x + 1) is a factor of top and bottom.

Bob

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 2012-11-30 02:19:20

Agnishom
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### Re: Indefinite integration

It is a factor of the denominator only.

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## #6 2012-11-30 03:26:55

bob bundy
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### Re: Indefinite integration

Sorry, you are right.  Selective blindness again.

I was having trouble with my minus signs.

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #7 2012-11-30 05:29:25

anonimnystefy
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### Re: Indefinite integration

The 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 2012-11-30 08:20:54

bobbym

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### Re: Indefinite integration

Hi;

Yes, it looks like a handmade one. That can be dangerous.

In mathematics, you don't understand things. You just get used to them.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #9 2012-11-30 14:38:17

noelevans
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### Re: Indefinite integration

Hi!

The third degree polynomial divided by the fifth degree polynomial can be written as an infinite
series as:

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!

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LaTex is like painting on many strips of paper and then stacking them to see what picture they make.