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integrating ln(X)

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    So I tyed in "integrate ln(x)" into wolphram alpha and I get

    xln(x)-x

    now the steps it gives are doing it by parts obviously

    U=ln(x) dV= dx

    dU = 1/x V= X

    since when can we include the dx at the end of every intergral with respect to x, as part of our integration ?

    can someone explain
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    ln(x)= ln(x)*1

    Use IBP: let u= ln(x), dv/dx=1.
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    (Original post by kingkongjaffa)
    So I tyed in "integrate ln(x)" into wolphram alpha and I get

    xln(x)-x

    now the steps it gives are doing it by parts obviously

    U=ln(x) dV= dx

    dU = 1/x V= X

    since when can we include the dx at the end of every intergral with respect to x, as part of our integration ?

    can someone explain

    I think you've assumed that because dv = dx, they have just taken the end "dx" and used that as v.

    in fact they have used the fact that f1mad has stated,
     \displaystyle \int ln(x)\ dx =  \displaystyle \int 1 \times ln(x)\ dx
    I can see that you know:
     \displaystyle \int uv\ dx
     =  \left[ uv \right] - \displaystyle\int v \frac{du}{dx}\ dx

    and then let  u equal the one that can be differentiated (ln(x)) and  v be the other (1)

    Back to your actual question :

    \frac{du}{dx} =1 and what has happened is that they've multiplied up by "dx" so that they have got du =dx.
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    (Original post by kingkongjaffa)
    So I tyed in "integrate ln(x)" into wolphram alpha and I get

    xln(x)-x

    now the steps it gives are doing it by parts obviously

    U=ln(x) dV= dx

    dU = 1/x V= X

    since when can we include the dx at the end of every intergral with respect to x, as part of our integration ?

    can someone explain
    See this example:

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    Thanks guys I understand now )

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