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Math Help - Ln Integration Problem

  1. #1
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    Ln Integration Problem

    \int \dfrac{(\ln x)^{3}}{2x}dx

    hint?
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    Re: Ln Integration Problem

    u = \ln{x}

    du = \frac{1}{x} \, dx

    substitute ...
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    Re: Ln Integration Problem

    \int \dfrac{(\ln x)^{3}}{2x}dx

    u = \ln x

    du = \dfrac{1}{x} dx

    \int \dfrac{(u)^{3}}{du}

    Next step?
    Last edited by Jason76; August 31st 2014 at 12:49 PM.
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    Re: Ln Integration Problem

    What? How did that "dx" get into the denominator? And what became of the "2" in the original integral?
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    Re: Ln Integration Problem

    Quote Originally Posted by HallsofIvy View Post
    What? How did that "dx" get into the denominator? And what became of the "2" in the original integral?
    How can I modify the "2" before putting a "du" in the integral?
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    Re: Ln Integration Problem

    \int \frac{(\ln{x})^3}{2x} \, dx = \frac{1}{2} \int (\ln{x})^3 \cdot \frac{1}{x} \, dx

    substitute ...

    \frac{1}{2} \int u^3 \, du
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    Re: Ln Integration Problem

    Quote Originally Posted by Jason76 View Post
    How can I modify the "2" before putting a "du" in the integral?
    Why would you want to "modify" a number? And why did you not answer my questions?
    Your original integral was \int \frac{(ln(x))^3}{2x}dx and skeeter suggested that you make the substitution u= ln(x) so that du= (1/x)dx.

    So \int\frac{(ln(x))^3}{2x}dx= \frac{1}{2}\int (ln(x))^3[(1/x)dx]= \frac{1}{2}\int u^3 du

    You should know that no such substitution will "move" dx into the denominator. There simply is no such thing as " \int \frac{f(u)}{du}"!
    Last edited by HallsofIvy; September 1st 2014 at 07:34 AM.
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    Re: Ln Integration Problem

    \int \dfrac{(\ln x)^{3}}{2x}dx

    u = \ln x

    \dfrac{1}{2}du = \dfrac{1}{x} dx - Some wrong turn is made here. ???

    \frac{1}{2} \int u^3 \, du

    \dfrac{1}{2} \dfrac{u^{4}}{4} + C

     \dfrac{u^{4}}{8} + C

    \dfrac{\ln(x)^{4}}{8} + C

    The book says the answer is 2(\ln x) + C, though ???
    Last edited by Jason76; September 2nd 2014 at 09:03 PM.
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    Re: Ln Integration Problem

    Quote Originally Posted by Jason76 View Post
    \int \dfrac{(\ln x)^{3}}{2x}dx

    u = \ln x

    \dfrac{1}{2}du = \dfrac{1}{x} dx - Some wrong turn is made here. ???
    NO! All that is happening is 1/2 is being taken out of the integral as a factor. It is NOT a part of the substitution!

    \frac{1}{2} \int u^3 \, du

    \dfrac{1}{2} \dfrac{u^{4}}{4} + C

     \dfrac{u^{4}}{8} + C

    \dfrac{\ln(x)^{4}}{8} + C

    The book says the answer is 2(\ln x) + C, though ???
    Well your book is clearly wrong. What do you get if you differentiate 2 ln(x) ? You should get 2/x, which is clearly not the same as your integrand...
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    Re: Ln Integration Problem

    So your saying my conclusion was correct, not the book's?

    Quote Originally Posted by Prove It View Post
    NO! All that is happening is 1/2 is being taken out of the integral as a factor. It is NOT a part of the substitution!



    Well your book is clearly wrong. What do you get if you differentiate 2 ln(x) ? You should get 2/x, which is clearly not the same as your integrand...
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  11. #11
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    Re: Ln Integration Problem

    Oh wait, I see what's going on. We would take out the 1/2 from the get go, and then do the substitution.

    So the conversions would look like this:

    u = \ln x

    du = \dfrac{1}{x}
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