Why does $\int_1^\sqrt2 \frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$ equal to $0$? In this question, the OP poses the following definite integral, which just happens to vanish:
$$\int_1^\sqrt2 \frac{1}{x}\ln\bigg(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\bigg)dx=0$$
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:
$$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t-sx^2+x^4}{tx-sx^2+x^3}\bigg)dx=0$$
 A: Put in other words we desire to show that:
$$I=\int_1^\sqrt2 \ln\left(\frac{2-2x^2+x^4}{2-2x+x^2}\right)\frac{dx}{x}=\int_1^\sqrt 2\frac{1}{x}\ln x dx=\frac18\ln^2 2$$
Let's take the LHS integral and split it in two parts.
$$I=\underbrace{\int_1^\sqrt2 \frac{\ln(2-2x^2+x^4)}{x}dx}_{\mathcal K}-\underbrace{\int_1^\sqrt2 \frac{\ln(2-2x+x^2)}{x}dx}_{\mathcal J}$$
$$\mathcal J=\int_1^\sqrt2 \frac{\ln(2-2x+x^2)}{x}dx\overset{\large x\to \frac{2}{x}}=\int_{\sqrt 2}^2 \frac{\ln(2-2x+x^2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx$$
$$\Rightarrow 2\mathcal J=\int_1^2\frac{\ln(2-2x+x^2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx$$
$$\mathcal J\overset{x\to x^2}=\int_1^\sqrt 2\frac{\ln(2-2x^2+x^4)}{x}dx-\frac{1}{8}\ln^22=\mathcal K-\frac{1}{8}\ln^2 2$$
$$\Rightarrow I=\frac{1}{8}\ln^2 2$$

Your conjecture is also correct, since by the same method we can show that:$$\int_1^{\sqrt{t}}\ln\left(\frac{t+sx^2+x^4}{t+sx+x^2}\right)\frac{dx}{x}=\int_1^\sqrt{t} \frac{\ln x}{x}dx=\frac{1}{8}\ln^2t$$
This time after we split the LHS integral into two parts, we will substitute in the second integral $\displaystyle{x\to\frac{t}{x}}$,
followed by an addition with the original $\mathcal J$ from there.
