How Can I Integrate a Ln Function Using Different Methods?

mjk1
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I have tried quite a few methods but haven't solved this problem, any help ?

\oint axLn(x/b) dx

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
 
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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?
 
u = ln(x/b) => du/dx = 1/xb (is that right ?)
dv = x dx => v = (x^2)/2

using formula...

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

how would you integrate this step ?
 
mjk1 said:
u = ln(x/b) => du/dx = 1/xb (is that right ?)
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.

dv = x dx => v = (x^2)/2

using formula...

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

how would you integrate this step ?
Surely you jest! "x2/x" is just x. You are asking about how to integrate x!
 
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 = \int^{ro}_{ri} [r^{2} - ro^{2} + \frac{ri^{2} - ro^{2}}{ln(ro/ri)}ln\frac{r}{ro}]rdr

where ro and ri are constants

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

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

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
 
nsequation.jpg
 
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