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Integrate a Ln function

  1. Oct 24, 2008 #1
    I have tried quite a few methods but havent solved this problem, any help ?

    [tex]\oint axLn(x/b) dx[/tex]

    where a and b are constants

    I have tried moving the 'a' outside the integration and solving xln(x/b) using integration by parts . I have also tried splitting xln(x/b) into xlnx -xlnb and integrating that but i'm still having problems.

    Any help would be appreciated
  2. jcsd
  3. Oct 24, 2008 #2


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    Looks to me like u= Ln(x/b), dv= x dx works nicely. Could you show where you have a problem with that integration by parts?
  4. Oct 24, 2008 #3
    u = ln(x/b) => du/dx = 1/xb (is that right ?)
    dv = x dx => v = (x^2)/2

    using formula.....

    [tex]\frac{x^2}{2}[/tex]Ln(x/b) - [tex]\int (x^2)/2xb[/tex]

    how would you integrate this step ?
  5. Oct 24, 2008 #4


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    No, it's not. It is [1/(x/b)] times the derivative of x/b or [b/x][1/b]= 1/x.
    A simpler way to do this is to use the fact that ln(x/b)= ln(x)- ln(b). Since ln(b) is a constant, it is clear that the derivative is 1/x, independent of b.

    Surely you jest! "x2/x" is just x. You are asking about how to integrate x!
  6. Oct 24, 2008 #5
    thanks for your help. this was a simplification of a Navier Stokes equation but I am having problem simplifying the equation after integration...

    original Integral

    Q = [tex]\int^{ro}_{ri} [r^{2} - ro^{2} + \frac{ri^{2} - ro^{2}}{ln(ro/ri)}ln\frac{r}{ro}]rdr[/tex]

    where ro and ri are constants

    so far I have managed to break it down to....

    [tex]\int^{ro}_{ri}[r^{2} - ro^{2}]rdr + \int^{ro}_{ri}[\frac{ri^{2} - ro^{2}}{ln(ro/ri)}ln\frac{r}{ro}]rdr[/tex]

    The first integral is straight foward but the second integral is a bit confusing

    can the second integral be written as
    arln(r/b) and solved ? where a and b are constants and then substitute the ri,ro for a and b after integration
  7. Oct 25, 2008 #6
    Last edited: Oct 25, 2008
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