help with problem

$\displaystyle y = {e^x}{ln{x}} $

Find

$\displaystyle a) \frac {dy}{dx} $

$\displaystyle b) \int_{e}^{e^2} \frac {1-ln{x}}{xln{x}} dx $

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- Jun 20th 2010, 11:49 PMBeachballOne differentiation question and one integration question.
help with problem

$\displaystyle y = {e^x}{ln{x}} $

Find

$\displaystyle a) \frac {dy}{dx} $

$\displaystyle b) \int_{e}^{e^2} \frac {1-ln{x}}{xln{x}} dx $ - Jun 20th 2010, 11:57 PMmr fantastic
- Jul 21st 2010, 05:41 PMBeachball
im still having trouble with it.. ive tried the substitution but i couldnt quite get it.. would someone be able to post the step by step result?

- Jul 21st 2010, 06:00 PMpickslides
Here's a kick off for you

$\displaystyle \int_{e}^{e^2} \frac {1-\ln{x}}{x\ln{x}} ~dx$

$\displaystyle u = \ln{x} \implies \frac{du}{dx}= \frac{1}{x}$

Now $\displaystyle \int\frac {1-\ln{x}}{x\ln{x}} ~dx = \int\frac{1}{x}\left(\frac {1-\ln{x}}{\ln{x}}\right) ~dx =\int \frac{1-u}{u}~du$ - Jul 21st 2010, 06:02 PMBeachball
yea ive got that far.. but just stuck and dont know where togo after that. :/

- Jul 21st 2010, 06:12 PMskeeter
$\displaystyle \frac{1-u}{u} = \frac{1}{u} - 1$

- Jul 21st 2010, 06:20 PMBeachball
thankyou all so much