How do I solve for the integral of 1/(xlnx) using logarithmic integration?

[tjbateh]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

From there I am stuck!

 

[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?

 

[tjbateh]

\frac{du}{u} ?

 

[rock.freak667]

\frac{du}{u} ?

yes so what is



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

 

[tjbateh]

is it just LN (x)??

 

[rock.freak667]

is it just LN (x)??

ln(u)

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

 

[tjbateh]

ln e²- ln e

 

[rock.freak667]

ln e²- ln e

No.

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

 

[tjbateh]

it is LN (LN(x))

 

[rock.freak667]

it is LN (LN(x))

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

 

[tjbateh]

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

 

[VietDao29]

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

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

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

 

[tjbateh]

it is 2

 

[tjbateh]

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

 

[VietDao29]

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

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

 

[tjbateh]

ok so that's the final answer??

 

[VietDao29]

ok so that's the final answer??

Yup. :)

 

[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!