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Integration how to start

  1. Apr 20, 2005 #1
    what is the integral of [tex]ln(1+2^x)[/tex]

    [tex]\int ln(1+2^x)=\frac{2^xln(2)}{1+2^x} [/tex]

    is this correct?
     
  2. jcsd
  3. Apr 20, 2005 #2
    Differentiate and see what you get.
     
  4. Apr 20, 2005 #3
    Try

    [tex] u = 2^x+1, du = 2^xln(2) [/tex]

    Then

    [tex] \int ln(1+2^x) dx = ln(2)\int (u-1)ln(u) du [/tex]

    Which can be done by parts.
     
  5. Apr 20, 2005 #4

    cepheid

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    That result you got looks suspiciously like the derivative of [itex] \ln(1+2^x) [/itex]! What I'm saying is, it looks like you differentiated using the chain rule:

    let u = 1 + 2^x

    [tex] \frac{d}{dx}[\ln(1+2^x)] = \frac{d}{dx}(\ln u) [/tex]

    [tex] = \frac{1}{u} \frac{du}{dx} [/tex]

    [tex] = \left(\frac{1}{1+2^x}\right) \frac{d}{dx}(1+2^x) [/tex]

    [tex] = \frac{(\ln2)2^x}{1+2^x} [/tex]

    But you were supposed to integrate! :smile:

    I just thought I'd point that out, so you could see the mistake.
     
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