# One differentiation question and one integration question.

#### Beachball

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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#### mr fantastic

MHF Hall of Fame
help with problem

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

Find

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

$$\displaystyle b) \int_{e}^{e^2} \frac {1-ln{x}}{xln{x}} dx$$
a) Use the product rule.

b) Substitute $$\displaystyle u = \ln x$$.

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

Beachball

#### Beachball

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

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

Beachball

#### Beachball

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

#### skeeter

MHF Helper
$$\displaystyle \frac{1-u}{u} = \frac{1}{u} - 1$$

Beachball

#### Beachball

thankyou all so much