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Math Help - How to integrate LN x over x.

  1. #1
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    How to integrate LN x over x.

    Hi. I am trying to solve dy/dx = yLnx/x, and to solve for y, I need to integrate Ln x over x. How can I do this?
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  2. #2
    Senior Member
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    Hello

    Quote Originally Posted by jaijay32 View Post
    Hi. I am trying to solve dy/dx = yLnx/x, and to solve for y, I need to integrate Ln x over x. How can I do this?

    Not sure if I get that right
    You want to solve

    \int \frac{ln(x)}{x}dx ?

    Using the product rule leads to ( u' = 1/x and v = ln(x) )

    \int \frac{ln(x)}{x}dx = ln(x) ln(x) - \int \frac{1}{x} ln(x) dx

    Add \int \frac{1}{x} ln(x) dx to both sides

    2 \int \frac{ln(x)}{x}dx = ln(x) ln(x)

    devide by 2

    \int \frac{ln(x)}{x}dx = 0.5 * ln(x) ln(x) + C

    Was it your question?

    Yours
    Rapha
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  3. #3
    Junior Member
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    Quote Originally Posted by jaijay32 View Post
    Hi. I am trying to solve dy/dx = yLnx/x, and to solve for y, I need to integrate Ln x over x. How can I do this?
    dy/dx = y{[ln x]/x}

    Rearranging,

    [1/y][dy]=[ln x]/[x] dx
    integrate both sides,
    ln y = integrate{[lnx]/x} dx
    ln y = 0.5(ln x)^2 +c

    Notice when you differentiate 0.5(ln x)^2 with respect to x you'll get back [ln x]/x.

    I hope I helped.
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