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 Post subject: Integrate 1/[x(x^2-2x-8)]Posted: Wed, 9 May 2012 14:14:28 UTC

Joined: Mon, 7 May 2012 14:05:46 UTC
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How can I integrate 1/{xsqrt[(x^2-2x-8]} without looking it up in a table or using a calculator? Correction!

Last edited by jsandifer on Wed, 9 May 2012 14:20:28 UTC, edited 1 time in total.

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 Post subject: Re: Integrate 1/[x(x^2-2x-8)]Posted: Wed, 9 May 2012 14:15:28 UTC
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jsandifer wrote:
How can I integrate 1/[x(x^2-2x-8)] without looking it up in a table or using a calculator?

Partial fractions, remembering to factorise x^2-2x-8 first.

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 Post subject: Re: Integrate 1/[xsqrt(x^2-2x-8)]Posted: Wed, 9 May 2012 15:41:45 UTC

Joined: Mon, 7 May 2012 14:05:46 UTC
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Note the correction. I forgot to include the square root.

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 Post subject: Re: Integrate 1/[xsqrt(x^2-2x-8)]Posted: Wed, 9 May 2012 16:40:21 UTC
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jsandifer wrote:
Note the correction. I forgot to include the square root.

Well, it depends on what you know. The easiest is to look at the meromorphic differential as a differential on the conic , and thus there is a local coordinate on such that with a meromorphic function of t.

If you carry that through, you get , or its reciprocal, is a substitution to try (in other words, you mapped the branch points of the square-root to 0 and infinity).

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 Post subject: Re: Integrate 1/[x(x^2-2x-8)]Posted: Sat, 12 May 2012 15:47:22 UTC

Joined: Mon, 7 May 2012 14:05:46 UTC
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Didn't work. Any other ideas?

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 Post subject: Re: Integrate 1/[x(x^2-2x-8)]Posted: Sat, 12 May 2012 15:49:57 UTC
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jsandifer wrote:
Didn't work. Any other ideas?

What do you mean by "didn't work"? It clearly does work.

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 Post subject: Re: Integrate 1/[x(x^2-2x-8)]Posted: Sat, 12 May 2012 18:09:12 UTC
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outermeasure wrote:
jsandifer wrote:
Didn't work. Any other ideas?

What do you mean by "didn't work"? It clearly does work.

My thoughts exactly.

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 Post subject: Re: Integrate 1/[x(x^2-2x-8)]Posted: Sat, 12 May 2012 22:46:33 UTC

Joined: Mon, 7 May 2012 14:05:46 UTC
Posts: 5
More specifically, if I use t=sqrt((x-4)/(x+2)) I get for an answer: Arctan(t/sqrt2)/sqrt(2). Wolfram Mathematica gives me: -Arctan((x+8)/(2sqrt(2)sqrt(x^2-2x-8))/(2sqrt(2)) and the answer I was given is: (5/3)Arcsec(abs(x-1)/3). Have you guys worked this out?

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 Post subject: Re: Integrate 1/[x(x^2-2x-8)]Posted: Sun, 13 May 2012 07:03:19 UTC

Joined: Sat, 26 Apr 2003 22:14:40 UTC
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Location: El Paso TX (USA)
This confirms what I thought all along. The problem should have been instead; then you end up with something like

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 Post subject: Re: Integrate 1/[x(x^2-2x-8)]Posted: Sun, 13 May 2012 07:57:12 UTC
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helmut wrote:
This confirms what I thought all along. The problem should have been instead; then you end up with something like

And what a difference that makes. Another point in favor of careful checking.

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 Post subject: Re: Integrate 1/[x(x^2-2x-8)]Posted: Tue, 15 May 2012 02:01:10 UTC

Joined: Mon, 7 May 2012 14:05:46 UTC
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Looks like everybody came out the same door, but I still wonder how Wolfram Mathematical got their answer.

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 Post subject: Re: Integrate 1/[x(x^2-2x-8)]Posted: Tue, 15 May 2012 03:51:16 UTC

Joined: Sat, 26 Apr 2003 22:14:40 UTC
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Location: El Paso TX (USA)
Mathematica does not like arcsec; all the arc functions are related anyway...

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