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Integral of 2/(y+1)?

  1. Feb 23, 2006 #1
    Hello,
    I need to take the integral of 2/(Y+1) over the interval [2,0]. The only analytical method I know is to take the anti-deriviative of the integrand and subtract the function values at the endpoints of the interval. I know that the anti-deriviate of (x)^-1 is ln(x), but I don't know how to do this problem without using the calculator (and I'd rather not do that). Can someone help?
     
  2. jcsd
  3. Feb 23, 2006 #2
    well y' = 2/(Y+1) remember that

    Integrate:
    2/(Y+1) * y'/y'
    2/y' * ln(y+1)
    substitute y'
    (y+1)ln(y+1)|2,0
    [3ln3 - 0]
    3ln3 is the answer
    hope i did it right...
     
  4. Feb 23, 2006 #3

    nrqed

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    well, clearly if you know that the antiderivative of [itex] 1/x [/itex] is [itex] ln (x) [/itex] then the antiderivative of [itex] 1/(x+1) [/itex] is [itex] ln (1 +x) [/itex] (check!! Take the derivative of the result!). and the antiderivative of [itex] 1/(Ax+B) [/itex] with A and B being constants is [itex] ln(A +B x) /A [/itex] (check!!).

    The way to prove it is to simply do the obvious change of variable, u = 1+x, dx=du and then integrate over u, which you know how to do. (If you do this change of variable to do yoru problem, don't forget to change the limits of integration.)
     
    Last edited: Feb 23, 2006
  5. Feb 23, 2006 #4
    Remember that dx=d(x+c), c=const.

    - Kamataat
     
  6. Feb 23, 2006 #5
    Hi guys,
    the correct answer is 2ln3, thanks for the help guys!
     
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