# Math Help - Help

1. ## Help

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

2. Using the quotient rule

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

Thus,

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

so

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

Not sure how this helps. Notice that

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

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

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

4. OK. I think I see where you want to go with this.

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

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

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

5. BRRB, many people look through these forums to learn something themselves. If you go back and delete all of the original questions, what is left makes no sense. There certainly is no sense in doing the addtional work of erasing the questions.

6. Originally Posted by HallsofIvy
BRRB, many people look through these forums to learn something themselves. If you go back and delete all of the original questions, what is left makes no sense. There certainly is no sense in doing the addtional work of erasing the questions.
Unfortunately, there is if you're getting help on things you're not meant to be getting help on (like work that counts towards your final grade) ....