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How to solve this integration Watch

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    (Original post by ElliotWalton)
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    Where did you stumble across this? Are you aware that there's no closed form?
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    (Original post by _gcx)
    Where did you stumble across this? Are you aware that there's no closed form?
    this is an inside interesting integrals
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    (Original post by ElliotWalton)
    this is an inside interesting integrals
    If you do want help with it. You'll need to know that,

    \displaystyle -\mathrm{Li}_2(-x) = \int \frac{\ln(x+1)}{x} \mathrm{d}x

    How can you rewrite \displaystyle \ln(x+1)\left(\frac{1}{x^2 + x}\right)?
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    (Original post by RDKGames)
    \displaystyle \int \frac{\ln(1+x)}{x^2+x} .dx = \int x^{-1}\cdot \frac{\ln(1+x)}{x} .dx then IBP
    can you write down all the workings?
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    (Original post by ElliotWalton)
    can you write down all the workings?
    Are you sure that what you write is correct?
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    (Original post by AnIndianGuy)
    He's right, trust me
    I have no ideas distinguishing whether He is correct or not.I hope you guys can show a specific working.THX
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    (Original post by RDKGames)
    \displaystyle \int \frac{\ln(1+x)}{x^2+x} .dx = \int x^{-1}\cdot \frac{\ln(1+x)}{x} .dx then IBP
    I was trying to get onto something different, ie.

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    Writing \displaystyle \int \frac{\ln(1+x)}{x^2 + x} \mathrm{d}x as \displaystyle \int \left(\frac{\ln(1+x)}{x} - \frac{\ln(1+x)}{x+1}\right) \mathrm{d}x would be a simpler approach imo.
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    (Original post by _gcx)
    I was trying to get onto something different, ie.

    Spoiler:
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    Writing \displaystyle \int \frac{\ln(1+x)}{x^2 + x} \mathrm{d}x as \displaystyle \int \left(\frac{\ln(1+x)}{x} - \frac{\ln(1+x)}{x+1}\right) \mathrm{d}x would be a simpler approach imo.


    I misread the Q when it came to the workings
 
 
 
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