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Annoying integral

  1. Jul 17, 2005 #1
    Hey guys i keep getting stuck with this:
    integral of ( ln(2x+1)dx)
    im supposed to use by parts
    heres what i have done
    u=ln(2x+1)
    du=2/2x+1
    dv=dx
    v=x

    then i apply the formula uv-vdu
    and i end up with another integral im supposed to use by parts for:
    integral of(2x/2x+1 dx)
    heres what i have done for that integral
    u=2x
    .5du=dx
    dv=1/2x+1
    v=ln(2x+1)

    then i apply the formula again, and since i have the same integal i stared wit, i add it to the lef side etc etc... but i dont get the correct answer...what does it look lie im doing wrong?
     
  2. jcsd
  3. Jul 17, 2005 #2

    dextercioby

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    Make a substitution first.

    [tex] 2x+1 = u [/tex]

    And then you'd have to integrate something proportional to [itex] \int \ln u \ du [/itex] which is really easy.

    Daniel.
     
  4. Jul 17, 2005 #3
    Math is easy once you accept the fact that substitution is the way to go ;)
     
  5. Jul 17, 2005 #4
    so make the u=2x+1 substitution in the first step of the problm?
     
  6. Jul 17, 2005 #5

    Pyrrhus

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    Yes as dexter noted

    [tex] \int \ln (2x+1) dx [/tex]

    [tex] u = 2x +1 [/tex]

    [tex] du = 2dx [/tex]

    [tex] \int \ln (u) \frac{du}{2} [/tex]
     
  7. Jul 17, 2005 #6
    gotcha, thanx.
     
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