While dealing with a definite integral on AoPS I discovered (I have to admit by pure chance) the following relation
$$\int_0^1\log\left(\frac{(x+1)(x+2)}{x+3}\right)\frac{\mathrm dx}{1+x}~=~0\tag1$$
The proof is quite easy, but feels kind of contrived. Indeed, just apply a self-similar substitution - $x\mapsto\frac{1-x}{1+x}$ - to the auxiliary integral $\mathcal I$ given by
$$\mathcal I~=~\int_0^1\log\left(\frac{x^2+2x+3}{(x+1)(x+2)}\right)\frac{\mathrm dx}{1+x}$$
And the result follows. However, to consider precisely this integral seems highly unnatural to me (in fact, as I mentioned earlier, this integral was just a by-product while evaluating something quite different and I discovered $(1)$ when experimenting with various substitutions).
The crucial point to notice concerning $\mathcal I$ is the invariance of the polynomial $f(x)=x^2+2x+3$ regarding the self-similar substitution which allows us to deduce $(1)$. Additionally for myself I am quite surprised by the special structure of $(1)$ since we have factors of the form $(x+1)$, $(x+2)$ and $(x+3)$ combined which calls for a generalization (although I found none yet).
It there a more elementary approach, not relying on such an "accident" like examining the integral $\mathcal I$ for proving $(1)$? Additionally, can this particular pattern be further generalized? Answers to both questions (also separately) are highly appreciated!
Thanks in advance!