Consider $$I_1 = \int_0^{\sqrt{2}} \ln\left(\dfrac{x^6+(n+6)x^4+(5n+6)x^2+2n}{x^6+(n+6)x^4+(4n+7)x^2+(4n-2)}\right)\,\mathrm dx$$ and $$I_2 = \int_{\sqrt{2}}^{\infty} \ln\left(\dfrac{x^6+(n+5)x^4+(4n+2)x^2}{x^6+(n+5)x^4+(3n+3)x^2+(2n-2)}\right)\,\mathrm dx$$
For $n\geq 2$, while checking from interent, I am getting $I_1+I_2<0$. Is there a way how to prove it? Is it true what I am observing?
We can make it formal if we write for each term $$x^6+ax^4+bx^2+c=(x^2-r_1)(x^2-r_2)(x^2-r_3)$$ So, expanding the logarithms, we face the summation of a series of integrals $$I_k=\int\log(x^2-r_k)\, dx=x \log \left(x^2-r_k\right)+2 \sqrt{r_k} \tanh ^{-1}\left(\frac{x}{\sqrt{r_k}}\right)-2 x$$ Now, it is just a matter of computing. For $n=1$, $I_1+I_2=0$ since the argument of the logarithm is equal to $1$. For $n>1$ (checked up to $n=10^5$), $(I_1+I_2)$ is always negative. $$\left( \begin{array}{cc} n & I_1+I_2 \\ 1 & 0 \\ 2 & -0.192471 \\ 3 & -0.249089 \\ 4 & -0.273505 \\ 5 & -0.285372 \\ 6 & -0.291233 \\ 7 & -0.293868 \\ 8 & -0.294640 \\ 9 & -0.294289 \\ 10 & -0.293238 \\ 20 & -0.273526 \\ 30 & -0.255691 \\ 40 & -0.241839 \\ 50 & -0.230857 \\ 60 & -0.221902 \\ 70 & -0.214419 \\ 80 & -0.208043 \\ 90 & -0.202521 \\ 100 & -0.197677 \\ 200 & -0.168542 \\ 300 & -0.153979 \\ 400 & -0.144786 \\ 500 & -0.138286 \\ 600 & -0.133368 \\ 700 & -0.129475 \\ 800 & -0.126291 \\ 900 & -0.123623 \\ 1000 & -0.121343 \\ 2000 & -0.108641 \\ 3000 & -0.102817 \\ 4000 & -0.099288 \\ 5000 & -0.096858 \\ 6000 & -0.095044\\ 7000 & -0.093629 \\ 8000 & -0.092484 \\ 9000 & -0.091532 \\ 10000 & -0.090725 \end{array} \right)$$ Obviously, $(I_1+I_2)$ shows a minimum value for $n=8$ and seems to tend to $0^-$ when $n \to \infty$.