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(lnx)/x watch

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    How would you integrate (lnx)/x ?
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    (Original post by Aired)
    How would you integrate (lnx)/x ?
    By parts:

    Let u = lnx
    .....dv/dx = 1/x

    and proceed as normal.
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    (Original post by V.P. Keys)
    By parts:

    Let u = lnx
    .....dv/dx = 1/x

    and proceed as normal.
    So... du/dx = 1/x and v = lnx.

    = (lnx)/x - ∫(lnx)/x

    :confused:
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    (Original post by Aired)
    So... du/dx = 1/x and v = lnx.

    = (lnx)/x - ∫(lnx)/x

    :confused:
    That's one way. Now rearrange the equation to make the integral you want the subject of it.
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    (Original post by Neapolitan)
    That's one way. Now rearrange the equation to make the integral you want the subject of it.
    Sorry I don't understand what you meant...
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    Take the integral on the RHS over to the LHS, so you get two lots of it.

    Incidentally, the first term on the RHS should be (lnx)2...
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    (Original post by Aired)
    Sorry I don't understand what you meant...
    ∫lnx/x dx = (lnx)² - ∫lnx/x dx

    So

    2 ∫lnx/x dx = (lnx)²

    ∫lnx/x dx = (lnx)²/2

    Plus a constant!
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    (Original post by Neapolitan)
    ∫lnx/x dx = (lnx)² - ∫lnx/x dx

    So

    2 ∫lnx/x dx = (lnx)²

    ∫lnx/x dx = (lnx)²/2

    Plus a constant!
    Oh! Thanks, I'm so stupid sometimes
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    (Original post by Aired)
    How would you integrate (lnx)/x ?
    Even easier.

    Let u = lnx
    => du/dx = 1/x
    => du = dx/x

    ∫(lnx)/x dx = ∫u du = ½u² = ½(lnx)² + c
 
 
 
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