<s>[INST] Evaluate $\int_0^1 \ln^2{\left(x^4+x^2+1\right)} \, \mathrm{d}x$ 
Evaluate $$\int_0^1 \ln^2{\left(x^4+x^2+1\right)} \, \mathrm{d}x$$

First thing I saw is $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ so the integral is the same as:
\begin{gather*}
\int_0^1\ln^2{\left(x^2+x+1\right)} \, \mathrm{d}x
+ \int_0^1\ln^2{\left(x^2-x+1\right)} \, \mathrm{d}x \\
+ 2\int_0^1\ln{\left(x^2+x+1\right)}\ln{\left(x^2-x+1\right)} \, \mathrm{d}x
\end{gather*}
I dont know if this helps though.  The last integral above seems to be the hardest, but still I don't even know how to evaluate first 2 integrals (perhaps Feynman's method)?
Source: https://tieba.baidu.com/p/4794735082
 [/INST] I would maybe try:
$$x^4+x^2+1=(x^2+\frac12)^2+\frac34$$
and so:
$$\int_0^1\ln^2(x^4+x^2+1)dx=\int_0^1\ln^2\left[(x^2+\frac12)^2+\frac{\sqrt{3}}{2}^2\right]dx$$
Now we know that:
$$\tan^2u+1=\sec^2u$$
so by letting:
$$\left[\frac{2}{\sqrt{3}}\left(x^2+\frac12\right)\right]=\tan(u)$$
$$\frac{4}{\sqrt{3}}xdx=\sec^2udu$$
we get:
$$I=\frac{\sqrt{3}}4\int\limits_{\frac\pi6}^{\frac\pi3}\ln^2\left[\frac32\tan^2u+\frac32\right]\left(\frac{\sqrt{3}\tan(u)-1}{2}\right)^{-1/2}\sec^2(u)du$$
and whilst the inside of the natural log simplifies nicely the rest is still rather ugly so I'm unsure if I can get a nice result from it but its worth a try :)
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