Proof that $\int \frac{1}{x}$ is $\ln(x)$

Question

When I was learning Calculus AB and Calculus II/III at my high school, I noticed that our textbooks never gave a full fundamental proof that $\int \frac{1}{x}$ is $\ln(x)$, and rather said that when you take the instantaneous derivatives of $\ln(x)$ the plotted points look like $\frac{1}{x}$

I don't really know how to start a proof for this. I'd like some help in creating the proof and understanding a proof for this equality, rather that approaching it from a graphical perspective.

Thanks in advance.

EDIT: I have searched this on the Internet and I'm just looking for an elementary proof that's easy to understand. Many of the proofs online are difficult for me to understand.

Answer 1

This question reveals a common misperception about mathematics: that mathematical statements are proved out of nothing, because they are simply "true". Nothing up my sleeves, then poof a theorem appears.

On the contrary, all theorems in mathematics must have a starting point. For some theorems, the starting point is naive "common knowledge" type stuff -- this makes it appear to be out of nothing, but it really isn't.

Natural log, exponent base e, these functions don't just exist on their own with all their properties. You define one of them, using one of the properties, and then use that definition to prove all of the other properties. You will never be able to prove all the properties, because you must have (at least) one to get started. You cannot build them out of "common knowledge" because their properties are too sophisticated for basic arithmetic, they depend on infinite limits and integrals and derivatives.

One starting point is to define $\ln x$ as $\int_1^x\frac{1}{t} dt$. If this is the definition of $\ln x$, then the OP is true by the fundamental theorem of calculus.

Another starting point is to define $e=\lim_{n\to\infty} (1+\frac{1}{n})^n$. Then, you define $e^x$ as an exponential, then prove properties about that function, then define $\ln x$ as its inverse.

A third starting point is to define $e^x=1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$, an infinite sum. Then, you define $e$ as the evaluation of this series at $x=1$, and proceed as in the previous version.

The reason all the commenters are asking what the definition is, is that in addition to these three there are probably half a dozen more ways your book could have begun the definitions. It would be a waste of time to guess and provide you a nice proof, that doesn't match the treatment given in your book.

Answer 2

In general, $y'(x)=\dfrac1{x'(y)}$ :

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$$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{f(x)-f(a)}{f^{-1}\Big(f(x)\Big)-f^{-1}\Big(f(a)\Big)}=\lim_{x\to a}\frac1{\dfrac{f^{-1}\Big(f(x)\Big)-f^{-1}\Big(f(a)\Big)}{f(x)-f(a)}}$$

$$=\frac1{\lim_{x\to a}\dfrac{f^{-1}\Big(f(x)\Big)-f^{-1}\Big(f(a)\Big)}{f(x)-f(a)}}=\frac1{\Big[f^{-1}\Big]'\Big(f(a)\Big)}$$

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Now, since $\ln x$ is the inverse of $e^x$, we have $(e^x)'=e^x~=>~\ln'x=\dfrac1{e^{\ln x}}=\dfrac1x$ . Furthermore, notice how the graphics of the exponential and logarithmic function are each other's reflection with regard to the first bisector, $f(x)=x$.

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Answer 3

From a little known definition of the logarithm*: $$\ln x=\lim_{n\to\infty}n(x^{\frac1n}-1),$$ $$\frac{d\ln x}{dx}=\frac{d}{dx}\lim_{n\to\infty}n(x^{\frac1n}-1)=\lim_{n\to\infty}\frac nnx^{\frac1n-1}=\frac1x.$$ *$x=(1+\frac yn)^n\iff y=n(x^{\frac1n}-1).$

Answer 4

Here's a fun proof that I made in this answer:

Consider the more basic power rule:

$$\int_1^xt^{n-1}\ dt=\frac{t^n}n\bigg|_1^x=\frac{x^n-1}n$$

Of course, we can't simply have $n=0$, but if we take the limit as $n\to0$, you end up with

$$\int_1^xt^{-1}\ dt=\lim_{n\to0}\frac{x^n-1}n$$

And since $x^n=e^{n\ln(x)}$, it follows that

$$\lim_{n\to0}\frac{x^n-1}n=\ln(x)\lim_{h\to0}\frac{e^h-1}h$$

But since we know that $\lim_{h\to0}\frac{e^h-1}h=1$, we end up with

$$\int_1^xt^{-1}\ dt=\ln(x)$$