A particular vanishing integral

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!

That's quite an impressive method to show that the integral vanishes.

For the first part I'll show using a different approach that your integral vanishes. $$\mathcal J=\int_0^1 \ln\left(\frac{x+3}{(x+2)(x+1)}\right)\frac{\mathrm dx}{x+1}\overset{x+1=t}= \color{blue}{\int_1^2\ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}}-\color{red}{\int_1^2 \frac{\ln t}{t}\mathrm dt}$$ Let's denote the blue integral as $\mathcal J_1$ then using the substitution $\frac{2}{t}\to t$ we get: $$\mathcal J_1=\int_1^2 \ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}=\int_1^2 \ln\left(\frac{2(t+1)}{t+2}\right)\frac{\mathrm dt}{t}$$ Adding both integrals from above gives us: $$\require{cancel} 2\mathcal J_1=\cancel{\int_1^2 \ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}}+\int_1^2 \frac{\ln 2}{t}\mathrm dt+\cancel{\int_1^2 \ln\left(\frac{t+1}{t+2}\right)\frac{\mathrm dt}{t}}=\ln^2 2$$ $$\Rightarrow \mathcal J_1=\color{blue}{\frac{\ln^2 2}{2}}\Rightarrow \mathcal J=\color{blue}{\frac{\ln^2 2}{2}}-\color{red}{\frac{\ln^2 2}{2}}=0$$

As for the second part, a small generalization outcomes by experimenting with the blue integral.

In particular, by the same approach we have: $$\int_1^a \ln\left(\frac{x+a}{x+1}\right)\frac{\mathrm dx}{x}=\int_1^a \frac{\ln x}{x}\mathrm dx$$ Which gives us a small generalization: $${\int_0^{a-1}\ln\left(\frac{x+a+1}{(x+1)(x+2)}\right)\frac{\mathrm dx}{x+1}=0}$$ Similarly, (with the substitution $\frac{ab}{x}\to x$) we get that: $$\int_a^b \ln\left(\frac{x+b}{x+a}\right)\frac{dx}{x}=\frac12 \ln^2 \left(\frac{b}{a}\right)=\int_a^b \ln\left(\frac{x}{a}\right)\frac{dx}{x}$$ And the following follows: $$\int_{a-1}^{b-1} \ln\left(\frac{a(x+b+1)}{(x+1)(x+a+1)}\right)\frac{dx}{x+1}=0$$ One might be interested in the following similar generalization too: $$\int_1^{t}\ln\left(\frac{x^4+sx^2+t^2}{x^3+sx^2+t^2x}\right)\frac{dx}{x}=0,\quad s\in R, t>1$$

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 $$ \int_0^1f(x)\frac{\mathrm{d}x}{x+a}=\int_0^1f\!\left(\tfrac{a(1-y)}{a+y}\right)\frac{\mathrm{d}y}{y+a}\tag3 $$