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    Hi,

    How the hell do I integrate

    (ln x)^n
    ???

    I though about setting u=lnx so it's the integral of u^n x e^u du, but I can't do the parts bit afterwards

    Cheers!
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    Use wolfram if you want to check your integrals
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    I can't really tell from that whether or not mine was right Maple agrees with me but I don't really trust it :p:
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    (Original post by vector)
    I can't really tell from that whether or not mine was right Maple agrees with me but I don't really trust it :p:
    You are right (ignoring +C )
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    Cool, cheers
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    Looks like a reduction formula question to me. Try doing parts with u=(lnx)^n, dv/dx = 1 and go nuts like you'd go nuts on ya **** around a purdy ladee.

    When you do this, setting your integral equal to I_n you get:

    I_n = nI_{n-1} + x(lnx)^n

    EDIT: Okay, so this requires you to be able to solve a first oder difference equation. I'm rusty with these badboys. I'll check it's do-able,

    EDIT 2: Yes. I'll spoilerise the method to solve the difference equation you get, for if you fancy having an atack yourself

    Spoiler:
    Show
    Multiply the equation through by 1/n!, and set S_n = \frac{I_n}{n!}. Then you'll have S_n = S_{n-1} + \frac{x(lnx)^n}{n!}. From here you can (hopefully) easily work out what S_n is, and thus what I_n is
 
 
 

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