Why is the integral of 1/x ln x?

At the end of the day, a lot depends on how you've defined e, I think.

You're right that integration is the reverse of differentiation. The integral of x^2 is x^3/3 because the derivative of x^3/3 is x^2.

But in the same way, can you look for something that differentiates to give 1/x? This is x^(-1) so using the same method as above, you may think that it's something involving x^0, but does this differentiate to give x^(-1)?

Since you haven't met logs or e yet, I'm not sure if you'll be able to get more of an explanation. I suggest either waiting until C3 or read a C3 textbook now / google it.

Yeah I thought x^0/0 which makes no sense because you can't divide by 0 and all that.. thats why I asked the teacher to elaborate but as you know here I am lol

I KIND OF met logs in FP1 Linear Laws thing but we haven't done it properly in C2 and we just winged the FP1 chapter by following laws of logs mindlessly... and indeed I have no idea what this mysterious irrational number e is ._.
I guess I will have to leave it until then yeah...

What, I thought e was just some irrational special number, I didn't know there were different e's?

There is a mysterious function ln(x) that you will meet in C3 that does have that property.

What, I thought e was just some irrational special number, I didn't know there were different e's?

Hello
We were doing FP1 topic on calculus and the teacher was about to do an example using 1/x but he was like oops it's an exception and the integral of 1/x is actually ln x.

I have no idea why this is true (my teacher didn't know/couldn't explain) and I have searched on Google but explanations there are incomprehensible for me.
Is there any way for me to understand this thing having only done C1 and half of FP1 so far? Will I be able to understand it by the end of A2 FM?


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Ok here is a 'basic i.e. non-formal' idea: the derivative of $e^x$ is itself. i.e. let $y=e^x$ then $\frac{dy}{dx}=e^x=y$

If we rearrange $\frac{dy}{dx}=y$ we see that $\frac{1}{y}dy=dx$ now integrating both sides gives $\int \frac{1}{y}dy=x+c$ where c is constant but recall $e^x=y \implies x=ln y$

so $\int \frac{1}{y}dy=ln(y)+c$

I don't see why ln x differentiates to 1/x in the first place and I don't know what log function is (Good start isn't it....). We sort of touched on logarithm as a part of FP1 chapter called Linear Law but we winged it by just using mysterious law of logs x.x

Ok I (kind of) am used to this log ab is log a+log b thing but uh...

And I assume you prove the derivative of e^x by using the derivative of ln x and using similar reasoning...

[Obviously, the involvement of 'e' is a massive spanner in the works at A-level, since it's a completely unfamiliar number and the definitions of it are pretty intractable ($\sum 1/n!$ is fairly easy to understand, but how the heck you work with the result is less so...)]

discussions of the graphs of 2^x, 3^x and their derivative graphs is a useful way to lead in to e... there should be a graph which coincides with its derivative graph.

For A-level students, you need to first learn implicit differentiation (I really mean the chain rule), then the above proof is the best you will get that you can understand.

On topic of $e^x$, one popular definition is via power series expansion:

$exp(x) = 1+ \frac{x^1}{1!} + \frac{x^2}{2!}+ \frac{x^3}{3!}+ ...$

Observe that it has the desired property that it (the RHS) differentiates to itself. In fact, it is easy to prove (you can try this), that this is the only (unique) power series that has this property - differentiate to itself.

You may notice, this definition however, has almost nothing to do with $e^x$, which is just a power function with a wierd number for the power base. This where we are going to have to stop the explanations, because it takes a serious amount of leg work (You may learn the truth if you go to university for Maths, by second year you may learn this), and accept, that:

$(1+ \frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+ ...)^x=e^x=exp(x) = 1+ \frac{x^1}{1!} + \frac{x^2}{2!}+ \frac{x^3}{3!}+ ...$

More specifically look at just:

$(1+ \frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+ ...)^x= 1+ \frac{x^1}{1!} + \frac{x^2}{2!}+ \frac{x^3}{3!}+ ...$

Clever factorization? Something obvious? No...Wish to know why? Time to apply for Maths at university.

For implicit differentiation, to find out why it is actually just an application of the chain rule, have a good think about why $\frac{d(y^2)}{dx}=2y\frac{dy}{dx}$.

It's helpful to remember what the equations mean. By the definition of natural growth dy = ydx, that's the whole point if e, it is an infinite series of self growth.So logically of you look at this from the inverse perspective, you need to scale out all that infinite self growth, which is dy/y = dx.So the function of the area of x^-1 has the exact same 'self limiting' effect, hence the equivalence.