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U = (1 + x) and integrate

  1. May 2, 2009 #1
    Hi!! I'm having trouble understanding my textbook, so can someone please explain to me how they got from


    [tex]\int1/(1+x)[/tex]2 dx, with the range of the integral from 0 to x/n

    to

    1 - n/(x+n)


    THank you So MuCh
     
  2. jcsd
  3. May 2, 2009 #2

    Pengwuino

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    Gold Member

    Re: integration

    Are you having trouble with the integration or the final answer?

    u substitution works fine here, let u = (1 + x) and integrate. The final answer is a simple algebraic trick, nothing spectacular.
     
  4. May 2, 2009 #3
    Re: integration

    I'm just having trouble with the integration ><. Oh wells
     
  5. May 2, 2009 #4

    Pengwuino

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    Gold Member

    Re: integration

    You're looking for the integral of [tex]\mathop \smallint \nolimits_0^{x/n} \frac{1}{{(1 + x')^2 }}dx'[/tex] (technically you're not allowed to have your variable of integration in your bounds)

    Make a substitution so that your integral now becomes [tex]\mathop \smallint \nolimits_0^{x/n} \frac{1}{{(u)^2 }}du\frac{{dx}}{{du}}[/tex] and remember to compute dx/du.

    What, when you take its derivative becomes [tex]\frac{1}{{u^2 }}[/tex]? Find what that is, substitute back in for what you had set u to and you can plug in your integration limits and wala!
     
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