One differentiation question and one integration question.

**Beachball** · June 20th 2010, 11:49 PM

help with problem

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

Find

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

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

**mr fantastic** · June 20th 2010, 11:57 PM

Originally Posted by Beachball

help with problem

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

Find

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

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

a) Use the product rule.

b) Substitute $u = \ln x$ .

If you need more help, please show all your work and say where you're stuck.

**Beachball** · July 21st 2010, 05:41 PM

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?

**pickslides** · July 21st 2010, 06:00 PM

Here's a kick off for you

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

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

Now $\int\frac {1-\ln{x}}{x\ln{x}} ~dx = \int\frac{1}{x}\left(\frac {1-\ln{x}}{\ln{x}}\right) ~dx =\int \frac{1-u}{u}~du$

**Beachball** · July 21st 2010, 06:02 PM

yea ive got that far.. but just stuck and dont know where togo after that. :/

**skeeter** · July 21st 2010, 06:12 PM

$\frac{1-u}{u} = \frac{1}{u} - 1$

**Beachball** · July 21st 2010, 06:20 PM

thankyou all so much

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