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Integrate 1/x^2+80x+1600

  1. Mar 21, 2015 #1

    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!


  2. jcsd
  3. Mar 21, 2015 #2
    to be clearer its 1/(x^2+80x+1600)dx!
  4. Mar 21, 2015 #3


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    Staff: Mentor

    I's start by factoring the denominator. What is the context of the question? Where did this integral come up? :smile:
  5. Mar 22, 2015 #4


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    To me, it looks like [itex]\frac{1}{(x+40)^{2}} [/itex]. So, if you set [itex]u=(x+40) [/itex], you get [itex] du=dx[/itex] and [itex]x=u-40 [/itex]. This leaves you with an eminently solvable integral.
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