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-   -   Integration, either I'm losing it or I've been working too long (https://www.askmehelpdesk.com/showthread.php?t=101953)

  • Jun 17, 2007, 08:27 AM
    albear
    integration, either I'm losing it or I've been working too long
    I need to integrate 1000(2t+1)^-2

    I thought it would go [(1000(2t+1)^-1)/-1 ] * t
  • Jun 17, 2007, 08:56 AM
    galactus
    Let

    This gives:

  • Jun 17, 2007, 09:15 AM
    albear
    sorry how do you apply dt=du/2 to get 500(integral)(1/u^2)du
  • Jun 17, 2007, 11:16 AM
    albear
    Hello
  • Jun 18, 2007, 01:39 PM
    ebaines
    You are trying to solve:



    correct?

    So replacing 2t+1 by u, and dt by du/2 gives you the form that Gallactus suggests. Now you should be able to integrate that form, and then the last step is to substitute 2t+1 for u to get the final answer.

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