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While dealing with a definite integral on AoPS I discovered (I have to admit by pure chance) the following relation

$$int_0^1logleft(frac(x+1)(x+2)x+3right)fracmathrm dx1+x~=~0tag1$$

The proof is quite easy, but feels kind of contrived. Indeed, just apply a self-similiar substitution - $xmapstofrac1-x1+x$ - to the auxiliary integral $mathcal I$ given by

$$mathcal I~=~int_0^1logleft(fracx^2+2x+3(x+1)(x+2)right)fracmathrm dx1+x$$

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 experimentating 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-similiar 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 $(1)$? Addionally, 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.
$$mathcal J=int_0^1 lnleft(fracx+3(x+2)(x+1)right)fracmathrm dxx+1oversetx+1=t= colorblueint_1^2lnleft(fract+2t+1right)fracmathrm dtt-colorredint_1^2 fracln ttmathrm dt$$
Let's denote the blue integral as $mathcal J_1$ then using the substitution $frac2tto t$ we get:
$$mathcal J_1=int_1^2 lnleft(fract+2t+1right)fracmathrm dtt=int_1^2 lnleft(frac2(t+1)t+2right)fracmathrm dtt$$
Adding both integrals from above gives us:
$$requirecancel 2mathcal J_1=cancelint_1^2 lnleft(fract+2t+1right)fracmathrm dtt+int_1^2 fracln 2tmathrm dt+cancelint_1^2 lnleft(fract+1t+2right)fracmathrm dtt=ln^2 2$$
$$Rightarrow mathcal J_1=colorbluefracln^2 22Rightarrow mathcal J=colorbluefracln^2 22-colorredfracln^2 22=0$$

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

In particular, by the same approach we have:
$$int_1^a lnleft(fracx+ax+1right)fracmathrm dxx=int_1^a fracln xxmathrm dx$$
Which gives us a small generalization:
$$int_0^a-1lnleft(fracx+a+1(x+1)(x+2)right)fracmathrm dxx+1=0$$
Similarly, (with the substitution $fracabxto x$) we get that:
$$int_a^b lnleft(fracx+bx+aright)fracdxx=frac12 ln^2 left(fracbaright)=int_a^b lnleft(fracxaright)fracdxx$$
And the following follows:
$$int_a-1^b-1 lnleft(fraca(x+b+1)(x+1)(x+a+1)right)fracdxx+1=0$$
One might be interested in the following similar generalization too:
$$int_1^tlnleft(fracx^4+sx^2+t^2x^3+sx^2+t^2xright)fracdxx=0,quad sin R, t>1$$

The Answer

I have used the substitution $(x+1)(y+1)=2$ before to good effect because
$$
int_0^1f(x),fracmathrmdxx+1=int_0^1f!left(tfrac1-y1+yright)fracmathrmdyy+1tag1
$$
If $f(x)=logleft(fracx+3(x+2)(x+1)right)$, then $f!left(frac1-y1+yright)=logleft(frac(y+2)(y+1)y+3right)$. Therefore
$$
int_0^1logleft(fracx+3(x+2)(x+1)right)fracmathrmdxx+1=int_0^1logleft(frac(y+2)(y+1)y+3right)fracmathrmdyy+1tag2
$$
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
$$
int_0^1f(x)fracmathrmdxx+a=int_0^1f!left(tfraca(1-y)a+yright)fracmathrmdyy+atag3
$$