Logarithmic Integration

1. Sep 21, 2009

tjbateh

1. The problem statement, all variables and given/known data

$$\int^{e^2}_{e}$$ $$\frac{1}{xlnx}$$ dx

2. Relevant equations

3. The attempt at a solution

I substituted U= xlnx
So DU= ($$\frac{1}{x}$$dx..........so Du * X = 1dx

From there I am stuck!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 21, 2009

rock.freak667

so if du =(1/x)dx where u=lnx

what does

$$\int \frac{1}{x lnx} dx$$

change to in terms of u and du?

3. Sep 21, 2009

tjbateh

$$\frac{du}{u}$$ ?

4. Sep 21, 2009

rock.freak667

yes so what is

$$\int \frac{1}{u} du = ?$$

5. Sep 21, 2009

tjbateh

is it just LN (x)??

6. Sep 21, 2009

rock.freak667

ln(u)

now since you are integrating from e2 to e, what is your integral equal to in terms of x?

7. Sep 21, 2009

tjbateh

ln e2- ln e

8. Sep 21, 2009

rock.freak667

No.

If you integrated 1/u du and got ln(u), and u = ln(x). What is ln(u) now?

9. Sep 21, 2009

tjbateh

it is LN (LN(x))

10. Sep 21, 2009

rock.freak667

right now so now compute Ln(ln(e2))-Ln(ln(e))

11. Sep 22, 2009

tjbateh

alright so LN (1/e^2)- LN (1/e) ???

12. Sep 22, 2009

VietDao29

No no, ln(ln(e2)) is definitely not ln(1/e2).

Now, let's do it step by step then. What is ln(e2)?

13. Sep 22, 2009

tjbateh

it is 2

14. Sep 22, 2009

tjbateh

and LN(e) is 1, so it would be LN (2)- LN (1)? Which is .693??

15. Sep 22, 2009

VietDao29

ln(1) = ln(e0) = 0, so, you can leave it as: ln(2) - ln(1) = ln(2). Taking an approximation is okay, though. :)

16. Sep 22, 2009

tjbateh

ok so that's the final answer??

17. Sep 22, 2009

VietDao29

Yup. :)

18. Sep 22, 2009

tjbateh

Great! Thanks for the help everyone! I just didn't think it made sense to have an LN in another LN, but I guess that works!! Thanks again!