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#1 2009-11-04 22:49:58
- Identity
- Member
- Registered: 2007-04-18
- Posts: 934
Integral
Can someone please show me the steps to solve:
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#2 2009-11-05 01:15:36
- TheDude
- Member
- Registered: 2007-10-23
- Posts: 361
Re: Integral
From Wikipedia (http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions):
The answer turns out to be surprisingly simple.
Wrap it in bacon
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#3 2009-11-05 02:15:11
- Identity
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- Registered: 2007-04-18
- Posts: 934
Re: Integral
Thanks TheDude, but I would like do know what method was used to solve it. I've tried integration by parts but it didn't work.
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#4 2009-11-05 04:48:53
- sarah12
- Guest
Re: Integral
I glad to know that you got the anwer to the problem that you Wanted.
#5 2009-11-05 08:35:13
- TheDude
- Member
- Registered: 2007-10-23
- Posts: 361
Re: Integral
Phew, ok, I finally worked this out. I'll hide it behind a button so the page doesn't explode.
Wrap it in bacon
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#6 2009-11-05 17:45:28
- Identity
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- Registered: 2007-04-18
- Posts: 934
Re: Integral
Wow! Thanks heaps TheDude, that was amazing!
Completing the square, trig substitution, exponent reduction, trig identities...wow
Last edited by Identity (2009-11-05 17:47:50)
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#7 2009-11-06 00:46:08
- TheDude
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- Registered: 2007-10-23
- Posts: 361
Re: Integral
Well, I want to emphasize that I just followed Wikipedia's proof step by step. They did the hard part, all I was left with was the algebra to generalize it.
Wrap it in bacon
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#8 2009-11-06 02:10:24
- Identity
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- Registered: 2007-04-18
- Posts: 934
Re: Integral
Still, much appreciated 
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#9 2009-11-06 05:08:40
- rzaidan
- Member
- Registered: 2009-08-13
- Posts: 59
Re: Integral
Hi Identity
In order to find this integral we do as the following:
∫1/(x ²+a ²) ^(3/2) dx = ∫(x ²+a ²)^-(3/2) dx
= ∫((x²(1+a ²* x^(-2))^(-(3/2)) dx by taking x ² az a common factor inside
=∫(x²)^-(3/2)(1+a ²* x^(-2))^-(3/2) dx
=∫x^(-3)(1+a ²* x^(-2))^-(3/2) dx then use substitution u = 1+a ²* x^(-2) to complete
please send a reply
Best Wishes
Riad Zaidan
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#10 2009-11-06 23:21:40
- Identity
- Member
- Registered: 2007-04-18
- Posts: 934
Re: Integral
Wow, thanks so much rzaidan, I worked through it and got the answer! Thanks again, very nice method
Last edited by Identity (2009-11-06 23:21:55)
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