# Integration x^-1

1. Apr 19, 2005

### no name

find
integration x^-1
i tried but reached nothing .....
plz explain

2. Apr 19, 2005

### Moo Of Doom

It's one of the ways to define ln(x). (Well, you use a definite integral for that, but it's practically the same).

$$\int_1^x{\frac{1}{t}dt}=\ln \left|x\right|$$

3. Apr 19, 2005

### dextercioby

You have to be kidding,right...?

$$\int \frac{dx}{x}$$

,or is it a joke...?

And what's this doing in the "Brain Teaser forum"...?April 1-st was a while ago...

Daniel.

4. Apr 19, 2005

### whozum

Dex you can be really mean sometimes.

5. Apr 19, 2005

### dextercioby

I have a mean looking avatar,don't I ? Why should i be gentle then ?

I appologize to the OP,whose name i don't know and couldn't possibly find out...

Daniel.

6. Apr 19, 2005

### whozum

Hehe, is that piccolo?

7. Apr 19, 2005

### dextercioby

No,he's a waiter...

Can the OP compute the derivative of $\ln x$ ...?It would be useful to compute the antiderivative...

Daniel.

8. Apr 19, 2005

### Galileo

No man, can't you see? That's Kami! Ofcourse he merged with Piccolo so now they are actually one. Good old Kami-sama was always pretty gentle actually. It was Piccolo who was mean.

9. Apr 19, 2005

### no name

could u prove it plz ?

10. Apr 19, 2005

### dextercioby

It can be shown starting with the definition that

$$\frac{d\ln x}{dx}=\frac{1}{x}$$.

Then

$$\int \frac{dx}{x}$$

becomes after the substitution

$$\frac{1}{x}=e^{t} \Rightarrow dx=-e^{-t} dt$$

$$\int e^{t}\left(-e^{-t} \ dt\right) =-\int dt=-t+C$$

Inverting the substitution,one finds

$$\int \frac{dx}{x}=-\ln\left(\frac{1}{\left|x\right|}\right) +C =\ln\left|x\right| +C$$

Daniel.