y= ln x / x

Find dy/dx

Hence show (integral sign with e^2 at top and e at bottom) 1-ln x / x ln x = ln 2 - 1

Results 1 to 6 of 6

- Jul 25th 2010, 06:19 AM #1

- Joined
- Jul 2010
- Posts
- 3

- Jul 25th 2010, 06:29 AM #2
Using the quotient rule

$\displaystyle \dfrac{dy}{dx} = \dfrac{\frac{1}{x} \cdot x - 1\cdot \ln x}{x^2}$.

Thus,

$\displaystyle \dfrac{d}{dx} \dfrac{\ln x}{x} = \dfrac{1 - \ln x}{x^2}$

so

$\displaystyle \displaystyle{\int}\dfrac{1 - \ln x}{x^2} = \dfrac{\ln x}{x} + c.$

Not sure how this helps. Notice that

$\displaystyle \displaystyle{\int}\dfrac{1 - \ln x}{x \ln x}\,dx = \int \dfrac{1}{x \ln x}\,dx - \int \dfrac{1}{x}\,dx$

The first can be integrated by letting $\displaystyle u = \ln x.$

- Jul 25th 2010, 06:33 AM #3

- Joined
- Jul 2010
- Posts
- 3

I got the first part, i just dont understand how the first part, ties in with the second part, because ive got to somehow get it in the form on f '(x)/ f(x) before i can integrate it

- Jul 25th 2010, 06:42 AM #4
OK. I think I see where you want to go with this.

$\displaystyle \displaystyle{\int} \dfrac{1 - \ln x}{x \ln x}\,dx = \displaystyle{\int}\frac{1 - \ln x}{x^2} \dfrac{1}{\frac{\ln x}{x}}\,dx$

Then let $\displaystyle u = \frac{\ln x}{x}$ so your integral becomes

$\displaystyle \displaystyle{\int} \dfrac{1}{u}\, du$.

- Jul 25th 2010, 09:24 AM #5

- Joined
- Apr 2005
- Posts
- 19,730
- Thanks
- 3011

- Jul 25th 2010, 12:43 PM #6