Integrate x^-1: Solutions & Explanation

[no name]

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

 

[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).

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

 

[dextercioby]

You have to be kidding,right...?

Are u talking about this

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

,or is it a joke...?

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

Daniel.

 

[whozum]

Dex you can be really mean sometimes.

 

[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.

 

[whozum]

Hehe, is that piccolo?

 

[dextercioby]

No,he's a waiter...

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

Daniel.

 

[Galileo]

Hehe, is that piccolo?

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.

 

[no name]

could u prove it please ?

 

[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.