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Integrate[ 2 x/(1 + 2 x), x]

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data

    When I plug in

    Integrate[ x/(1 + x), x]

    into Mathematica, I get

    x - Log[1 + x]

    which I've been able to recreate on paper. However, when I plug in

    Integrate[ 2 x/(1 + 2 x), x]

    I get

    1/2 + x - 1/2 Log[1 + 2 x]

    but on paper I get

    x - 1/2 Log[1 + 2 x]

    My question is, where does the constant (1/2) come from? Is it me or Mathematica that's wrong?
     
  2. jcsd
  3. Apr 20, 2009 #2
    Integrate[ x/(1 + x), x] =

    x - 1/2 Log[1 + 2 x] + 1/2
    or
    x - 1/2 Log[1 + 2 x]
    or
    x - 1/2 Log[1 + 2 x] + c

    What's the difference?
     
  4. Apr 20, 2009 #3
    5 or 10 points on a test :) depending on how a teacher would grade such mistakes. I'm using Mathematica to check my work and these kinds of things make me wonder if my procedure was off or not.
     
  5. Apr 20, 2009 #4
    I remember asking same question like two years ago. I don't think there's any difference. Here's the matlab:

    >> syms x;
    >> int( 2*x/(1 + 2*x), x)
    ans =
    x-1/2*log(1+2*x)

    But, regardless of what constant you get, in the end there's always "c"
    one constant+1/2 = another constant ('c'). The question doesn't have one single answer.

    (Also, I don't know why Mathematica is giving 1/2 in the end)
     
  6. Apr 20, 2009 #5
    You guys rely WAAAY too much on machines for simple calculations. Back in the native American reservation, I only had small beads for calculus.

    Anyway, you are doing it wrong, Mathematica is right.

    Int[2x/(1+2x)] dx
    Substitute u=1+2x so that dx=du/2
    Notice that u-1=2x

    Therefore, the integral is
    Int[(u-1)/2u] du
    which, by splitting it up, is
    Int[1/2] du +Int[-1/2u]
    which is
    u/2 -1/2 ln(u)+c
    Don't forget to put it in terms of x

    (1+2x)/2-1/2ln(1+2x)+c

    This is probably where you messed up you thought (1+2x)/2 is x.
    Well, it is actually,

    1/2+x-1/2ln(1+2x)+c
     
  7. Apr 20, 2009 #6
    I wasn't thinking about substitution then but here's how he/me were doing (I think):

    =int (2x / (1+2x))
    = int (1-1/(1+2x))
    =x - 1/2.ln(1+2x) + c

    I wouldn't call missing 1/2 wrong if you are using partial fractions (not substitution).
     
  8. Apr 20, 2009 #7
    *Slaps hand on face* forgot that the +c made the 1/2 completely meaningless. Sorry, that's a habit of mine.

    Of course, x - 1/2 Log[1 + 2 x] + c is the simplified solution and your professor may take off some points if you hand in my/Mathematica's answer.
     
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