# Integrate a Ln function

1. Oct 24, 2008

### mjk1

I have tried quite a few methods but havent 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

2. Oct 24, 2008

### HallsofIvy

Staff Emeritus
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?

3. Oct 24, 2008

### mjk1

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 ?

4. Oct 24, 2008

### HallsofIvy

Staff Emeritus
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!

5. Oct 24, 2008

### mjk1

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

6. Oct 25, 2008

### mjk1

Last edited: Oct 25, 2008