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