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    How do you partially integrate the following with respect to x?

    (1/x)*ln(xe^y) dx

    I used integration by parts to get:

    ln(xe^y)lnx - ((lnxe^y)^2)/2) + c

    But apparently the answer is just ((lnxe^y)^2)/2) + c?

    how does the ln(xe^y)lnx cancel out?

    thanks
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    Well one thought is that \ln x e^y = \ln x + \ln e^y= \ln x +y BTW, my answer is nothing like yours either, have I written the Q out incorrectly.? Here it is in tex \int \frac{ln(xe^y)}{x}\;dx
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    (Original post by zomgleh)
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    You don't need partial integration, the integrand is already in the form y'y. Anyway, you made a mistake when differentiating.
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    (Original post by zomgleh)
    How do you partially integrate the following with respect to x?

    (1/x)*ln(xe^y) dx

    I used integration by parts to get:

    ln(xe^y)lnx - ((lnxe^y)^2)/2) + c

    But apparently the answer is just ((lnxe^y)^2)/2) + c?

    how does the ln(xe^y)lnx cancel out?

    thanks
    \int \frac{1}{x}ln(xe^y) dx=\int e^y\frac{1}{xe^y}ln(xe^y) dx=
    =\frac{ln^2(xe^y)}{2}+C
    using
    \frac{dln(ax)}{dx}=\frac{1}{ax}a
    and so
    \int f'(x)\frac{1}{f(x)}ln[f(x)] dx=\frac{ln^2(f(x))}{2}+C
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    (Original post by Lord of the Flies)
    You don't need partial integration, ...
    Can you expand on that? The integral would be +f(y) rather than +C with p.i. I don't understand why you said "you don't need PI" surely that is merely irrelevant to the integration but the "answer" in the case of pi requires +f(y), so he does "need" it in the context of the question.
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    (Original post by nerak99)
    Can you expand on that? The integral would be +f(y) rather than +C with p.i. I don't understand why you said "you don't need PI" surely that is merely irrelevant to the integration but the "answer" in the case of pi requires +f(y), so he does "need" it in the context of the question.
    Not sure what what you're talking about, I merely pointed out that there is no need to partially integrate by noticing that:

    \displaystyle\int \frac{\ln x e^y}{x}\,dx=\int \ln xe^y d\ln xe^y
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    I think I know whats going on here, the OP meant integrate "partially", i.e partial integration where x and y may both be functions of another variable, not by parts. Consequently, at the end, instead of adding C we have to add f(y).

    I think you just meant that he doesn't have to use integration by parts. I think the problem is a bit artificial in the sens that it was constructed from a partial differentiation and then not simplified in order to create the problem.
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    (Original post by nerak99)
    I think I know whats going on here, the OP meant integrate "partially", i.e partial integration where x and y may both be functions of another variable, not by parts. Consequently, at the end, instead of adding C we have to add f(y).
    I have always seen the term partial integration as meaning integration by parts, and Google seems to agree - but perhaps it can also refer to what you mention. However, given that the OP said "the answer is ln(xe^y)^2/2 + C", and that they also used the term "by parts", I assume they meant integration by parts.
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    thanks for all your inputs, found the answer so thought of sharing it--

    break it down into (1/x lnx + y/x) and integrate with respect to x.

    and I meant partial integration, as in treat the y (and any other variable) as a constant, which is not to be confused with integrating by parts (which is something I used to integrate the function without simplifying it)

    cheers
 
 
 
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