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An easy integral

  1. May 10, 2010 #1
    An "easy" integral

    1. The problem statement, all variables and given/known data

    solve ∫x/(x+1)


    3. The attempt at a solution

    let u = x + 1 ,therefore, x = u - 1

    hence, ∫x/(x+1) = ∫(u-1)/u = ∫u/u - ∫1/u = ∫1 - ∫1/u = u - lnu = x + 1 - ln(x+1)
    =x+1+ln(1/(x+1))

    differentiating this gives me x + 2 which does not make sense!
    Please help!
     
  2. jcsd
  3. May 10, 2010 #2

    Cyosis

    User Avatar
    Homework Helper

    Re: An "easy" integral

    Use the chain rule.
     
  4. May 10, 2010 #3

    Dick

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    Science Advisor
    Homework Helper

    Re: An "easy" integral

    Differentiating that does NOT give you x+2. Try differentiating again. Carefully, this time.
     
  5. May 10, 2010 #4

    Mark44

    Staff: Mentor

    Re: An "easy" integral

    Why not leave it as x + 1 - ln(x + 1)? I don't see any advantage in rewriting the last term as ln(1/(x + 1)), and it makes taking the derivative more complicated.

    Also, don't forget the constant of integration!
     
  6. May 10, 2010 #5
    Re: An "easy" integral

    So you agree that x + 1 - ln(x+1) = x+1+ln(1/(x+1)) ?

    what would be the steps for differentiating x+1+ln(1/(x+1)) without changing it back to : x + 1 - ln(x+1) ?

    Thanks!
     
  7. May 10, 2010 #6
    Re: An "easy" integral

    Oh ok thanks, doesn't matter, i got it!

    (At night I can think clearer than in the daytime!)
     
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