Fake proof that $0=1$ using $\int\frac{1}{x\ln x}\,\mathrm{d}x$ and integration by parts

$\newcommand{\d}{\mathrm{d}}$I saw, in a series of joke proofs that Santa Claus exists, the following:

We want to evaluate the indefinite integral of: $$\int\color{green}{\frac{1}{\ln x}}\cdot\color{red}{\frac{\d x}{x}}$$

Using the standard IBP, I consider $\color{green}{u}$ and $\color{red}{\d v}$:

$$\int\color{green}{\frac{1}{\ln x}}\cdot\color{red}{\frac{\d x}{x}}=\color{green}{\frac{1}{\ln x}}\cdot\color{red}{\ln x}-\int\color{red}{\ln x}\color{green}{\left(-\frac{1}{x\cdot\ln^2x}\right)\,\d x}=1+\int\frac{1}{\ln x}\cdot\frac{\d x}{x}$$

From which we "conclude" that $0=1$ by subtracting the indefinite integral from both sides.

A simple substitution concludes that in fact the indefinite integral is $\ln\ln x$, but I'm really curious where the fallacy is here; I think I've found it, but would like confirmation. I do not see any division by zero errors, the chain rule is correctly applied.

Going back to the derivation of IBP, the fallacy possibly lies in the fact that $u v$ is a constant function with zero derivative, so I need to be careful with constants of integration, i.e. include $C=-1$.

Is that right? And in general, if $uv$ is a constant function, will IBP create silly expressions like this?

An indefinite integral denotes the set of all antiderivatives of its integrand. What you've proven is$$\left\{f|f^\prime=\tfrac{1}{x\ln x}\right\}=\{1+g|g^\prime=\tfrac{1}{x\ln x}\}.$$That these two sets of functions are the same implies a function is an element of one iff it's an element of the other, but doesn't say which other elements the second set has.