TITLE: A particular vanishing integral
QUESTION [16 upvotes]: 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!

REPLY [7 votes]: The Answer
I have used the substitution $(x+1)(y+1)=2$ before to good effect because
$$
\int_0^1f(x)\,\frac{\mathrm{d}x}{x+1}=\int_0^1f\!\left(\tfrac{1-y}{1+y}\right)\frac{\mathrm{d}y}{y+1}\tag1
$$
If $f(x)=\log\left(\frac{x+3}{(x+2)(x+1)}\right)$, then $f\!\left(\frac{1-y}{1+y}\right)=\log\left(\frac{(y+2)(y+1)}{y+3}\right)$. Therefore
$$
\int_0^1\log\left(\frac{x+3}{(x+2)(x+1)}\right)\frac{\mathrm{d}x}{x+1}=\int_0^1\log\left(\frac{(y+2)(y+1)}{y+3}\right)\frac{\mathrm{d}y}{y+1}\tag2
$$
and since the two sides of $(2)$ are negatives, they are both $0$.

A Generalization
We can generalize $(1)$ by letting $(x+a)(y+a)=a(1+a)$, then we get
$$
\int_0^1f(x)\frac{\mathrm{d}x}{x+a}=\int_0^1f\!\left(\tfrac{a(1-y)}{a+y}\right)\frac{\mathrm{d}y}{y+a}\tag3
$$