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Is there such a thing a literaly integrating ln x?

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    Hi there

    Is there such a thing a literaly integrating ln x?
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    Yes, it is x\ln x - x.

    You can find it using integration by parts.
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    Yup. choose, as JB says, parts, using Ln(x) = Ln(x) x (1) - derive the Log part, integrate 1 for the other
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    How do you figuratively integrate a function? :confused:
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    (Original post by jackie11)
    Hi there

    Is there such a thing a literaly integrating ln x?
    I'm gonna take a guess and say you've got a C3 exam 2morrow.
    You should avoid integrating Ln X, that's why in integration by parts you make u=Ln x no matter what that way you only have to differentiate it and not integrate it.
    This method works because i tried it with a question and came out with the right answer.
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    Here is an example from my book:

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    (Original post by EEngWillow)
    How do you figuratively integrate a function? :confused:
    Eh?

    NO!
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    (Original post by Hasufel)
    Eh?

    NO!
    well your question said "literally" and dont really know what you mean by using that word in this context?
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    xln(x) - x + c
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    It could be a misspelling of "illiterately".
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    On a more serious note - the posters who gave the answer are correct, but you should try it yourself. Integrate by parts, using the trick u = \ln x, \frac{dv}{dx} = 1. It may seem weird, but it works.

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