How Do You Integrate 1/(x^2+80x+1600)?

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Daniel Tyler
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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=0

Any help is much appreciated!

Regards

Daniel
 
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to be clearer its 1/(x^2+80x+1600)dx!
 
Daniel Tyler said:
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=0

Any help is much appreciated!

Regards

Daniel

Daniel Tyler said:
to be clearer its 1/(x^2+80x+1600)dx!

I's start by factoring the denominator. What is the context of the question? Where did this integral come up? :smile:
 
Daniel Tyler said:
to be clearer its 1/(x^2+80x+1600)dx!
To me, it looks like \frac{1}{(x+40)^{2}}. So, if you set u=(x+40), you get du=dx and x=u-40. This leaves you with an eminently solvable integral.
 
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