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danny740
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Join Date: Nov 2014
Posts: 13

Integrate 1/x^2+80x+1600

03/21/2015 4:38 PM

Hi,
I'm v rusty on my integration so was wondering could anyone give me some guidance on how to find the solution to the following integral
integral (1/x^2+80x+1600)dx between the limits x=80 and x=0Any help is much appreciated!
RegardsDaniel
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danny740
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#1

Re: Integrate 1/x^2+80x+1600

03/21/2015 4:40 PM

to be clearer its 1/(x^2+80x+1600)dx!
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Rixter
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#4
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Re: Integrate 1/x^2+80x+1600

03/21/2015 8:13 PM

Hint1: Try factoring the denominator

Hint2: Substitution of variable
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Dan A
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#6
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Re: Integrate 1/x^2+80x+1600

03/22/2015 3:26 AM

1/(x^2+80x+1600) = 1/(x+40)^2

Let u = x+40

then du = dx

problem then becomes: integral of 1/u^2 = -1/u =-1/(x+40)
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SolarEagle
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#3

Re: Integrate 1/x^2+80x+1600

03/21/2015 5:15 PM

http://www.freemathhelp.com/forum/archive/index.php/t-44008.html?s=81349e7c156e89af0e48100743b7ef6d

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Just an Engineer
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#5

Re: Integrate 1/x^2+80x+1600

03/22/2015 3:23 AM

hint:

1/x^2 is really x^-2

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