Csdrhrt

In this question, the OP poses the following definite integral, which just happens to vanish:
$$int_1^sqrt2 frac{1}{x}lnbigg(frac{2-2x^2+x^4}{2x-2x^2+x^3}bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=sqrt[3]{2}$ to $x=sqrt{2}$.

This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.

Any ideas?

EDIT: I believe that this more general integral also vanishes:
$$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{tx-sx^2+x^3}bigg)dx=0$$

We can indeed prove the result by symmetry. Put in other words we desire to show that: $$I=int_1^sqrt2 frac{1}{x}lnbigg(frac{x^4-2x^2+2}{x^2-2x+2}bigg)dx=color{green}{int_1^sqrt 2frac{1}{x}ln x dx=frac18ln^2 2}$$
First let us take the LHS integral and split it in two parts:
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt2 frac{ln(x^2-2x+2)}{x}dx}$$
For the second one $(I_2)$ we will substitute $displaystyle{x=frac{2}{t}rightarrow dx=-frac{2}{t^2}dt}$:
$$I_2=int_{sqrt 2}^2 frac{lnleft(frac{2(t^2-2t+2)}{t^2}right)}{frac{2}{t}}frac{2}{t^2}dtoverset{t=x}=int_{sqrt 2}^2 frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx$$
Adding with the original $I_2$ leads to:
$${2I_2=int_1^2frac{ln(x^2-2x+2)}{x}dx+int_{sqrt 2}^2frac{ln 2 -2ln x}{x}dx}$$
$${Rightarrow I_2=frac12int_1^2 frac{ln(x^2-2x+2)}{x}dx-frac{1}{8}ln^22}overset{x=t^2}=color{blue}{int_1^sqrt{2}frac{ln(t^4-2t^2+2)}{t}dt-frac{1}{8}ln^2 2}$$
$$I=color{red}{int_1^sqrt2 frac{ln(x^4-2x^2+2)}{x}dx}-color{blue}{int_1^sqrt{2}frac{ln(x^4-2x^2+2)}{x}dx+frac{1}{8}ln^2 2}=color{green}{frac18ln^2 2}$$

Your conjecture is indeed also correct since by the exact same method we can show that: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{t-sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx=frac{1}{8}ln^2t$$
And this time after we split the LHS integral into two parts, we will substitute in the second integral $displaystyle{x=frac{t}{y}}$ ($t$ is a constant here),
followed by an addition with the original $I_2$ from there and the result follows. Of course it is valid in the following form too: $$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t+sx^2+x^4}{t+sx+x^2}bigg)dx=int_1^sqrt{t} frac{1}{x}ln xdx$$