system
Given the following post, respond as if you're commenting on Reddit:
human
Why is the integral of 1/x = ln(x)+C?

I mean I get that you cannot integrate it normaly as in:
x-1 --> (x0 )/0
because you obviously cannot divide by zero.
So why does this just happen to equal ln(x)?
gpt
First of all, you mean to ask why the *antiderivative* of 1/x is ln(x) + C. An *integral* is a limit of Riemann sums, i.e., a single number, and it has various interpretations: area, volume, probability, etc. The fundamental theorem of calculus simply provides a link between integrals and antiderivatives. But the two objects are not the same thing.

Many introductory textbooks simply define ln(x) to be an antiderivative of 1/x, and that's that. So the answer to your question is "because it is". A better way of defining ln(x) from first principles, at least in real analysis, is that g(x) = ln(x) is the inverse function of f(x) = e^(x). The immediate question is then whether these two definitions are equivalent. They are.

Consider the function y = g(x) = ln(x). By definition, e^(y) = x. Now use implicit differentiation (w.r.t. *x*) to find that

> e^(y) (dy/dx) = 1

Solve for (dy/dx) to get

> dy/dx = 1/e^(y)

But note that e^(y) = x, so we get

> dy/dx = 1/x

Voila. If we define ln(x) to be the inverse of e^(x), we still get that its derivative is 1/x. (By a similar method, you can show that if we define ln(x) to be a function whose derivative is 1/x, then ln(x) is just the inverse function of e^(x).)