Hello, /sci/. I'm wondering how to resolve this discrepancy with the indefinite integral of 1/x.

$\int_ \frac{1}{x} dx = ln(x) + C$

So this is cool and accepted, right? However...

$\frac{d}{dx} ln(x) = \frac{1}{x}$

And

$\frac{d}{dx} ln(3x) = \frac{1}{3x} \cdot 3 = \frac{1}{x}$

So, my question is: how come

$\int_ \frac{1}{x} dx \neq ln(3x) + C$

since $\frac{1}{x}$ is a valid antiderivative of $ln(3x)$? This question would apply for any real constant multiple of x inside the argument of the natural log function, I believe, not just 3.