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Math Help - Is this a legitimate way of finding the integral of lnx?

  1. #1
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    Is this a legitimate way of finding the integral of lnx?

    \int{\ln{x} \ dx}

    = \int{\ln{x}(1) \ dx} //pull out a '1'

    = \ln{x}(x) - \int{\frac{1}{x}x} //use integration by parts

    = x\ln{x} - \int{1}

    = x\ln{x} - x

    = x(\ln{x} - 1)

    Is pulling out a 1 in the first step a legitimate method when using integration by parts, or is it just a coincidence that it works? Thanks!
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  2. #2
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    Pull out a 1? There is a 1 in every equation since multiplying by 1 does nothing.

    Just doing integration by parts will work.
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  3. #3
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    I think my understanding of integration by parts isn't too good.

    I'm not sure how to perform it without there being two functions to work with, which is why I pulled out the one. What's the proper way of dealing with this?
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  4. #4
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    If we have an x, there is always an implied 1. There is a one in every function you can think since multiplying by a 1 does nothing to an equation.
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  5. #5
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    So my method was right, but you don't think of it as pulling out a one? I just did that to help my head get around it.
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  6. #6
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    Why pull out something that already exists?
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  7. #7
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    Is this what you're doing? If it's, then it's valid, of course:

    \displaystyle \int\ln{x}\;{dx} = \int(\ln{x})({1})\;{dx} = \int(\ln{x})(x)'\;{dx}

    \displaystyle = (\ln{x})(x)-\int\left(\ln{x}\right)'\cdot x\;{dx} = (\ln{x})(x)-\int (1)\;{dx}

     \displaystyle = x\ln{x}-x+k = x\left(\ln{x}-1\right)+k.
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  8. #8
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    Yeah, that's what I'm doing. Thanks.
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