<s>[INST] Integral $\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$ A while ago I encountered this integral $$I=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$$ To be fair I spent some time with it and solved it in a  heuristic way, I want to avoid that way so I won't show that approach, but the result I got is $\frac{\pi}{6} \ln(2+\sqrt 3)$  so I bet it can be shown in a nice way.
Solving other integrals I also encountered this one: $$J=\int_0^\infty \frac{\ln(1+x^2+x^4)}{1+x^2}dx=\pi \ln(2+\sqrt 3)$$ Which is pretty easy to compute, so most of my time I tried to show that $J=6I$, however without explictly evaluating them I had no luck.
Also I tried to use partial fractions: $$I=\frac12 \left(\int_0^1 \frac{\ln(1+x)}{x^2+\sqrt 3x +1}dx - \int_0^1 \frac{\ln(1+x)}{x^2-\sqrt 3 x+1}dx\right) $$
Considering: $$K(t) =\int_0^1 \frac{\ln(1+x)}{x^2-2\cos(t)x+1}dx$$
We have that $I=\frac12 \left(K\left(\frac{5\pi}{6}\right)-K\left(\frac{\pi}{6}\right)\right) $ and since: $$\frac{\sin t}{x^2-2x\cos t+1}=\frac{1}{2i}\left(\frac{e^{it}}{1-xe^{it}}-\frac{e^{-it}}{1-xe^{-it}}\right)=\Im \left(\frac{e^{it}}{1-xe^{it}}\right)=$$
$$=\sum_{n=0}^{\infty} \Im\left(x^n e^{i(n+1)t}\right)=\sum_{n=0}^\infty x^n\sin((n+1)t)$$
$$\small \Rightarrow K(t)=\frac12 \left(\frac{1}{\sin \left(\frac{5\pi}{6}\right)}\sum_{n=0}^\infty  \sin\left(\frac{5\pi}{6} (n+1)t \right) + \frac{1}{\sin \left(\frac{\pi}{6}\right)}\sum_{n=0}^\infty  \sin\left(\frac{\pi}{6} (n+1)t \right)\right)\int_0^1 x^n \ln(1+x)dx$$
$$=2\sum_{n=0}^\infty  \sin\left(\frac{\pi}{6} (n+1)t \right)\int_0^1 x^n \ln(1+x)dx$$
Now I don't know how to deal with the integral and the sum combined. 
 [/INST] I thought about the comment you left the other day, and it has led me to this alternative approach, which is more direct than my other answer. Thanks a lot, I have learned so much from you!
Let $\omega=\sqrt[3]{-1}$, and let Ti2 denote the inverse tangent integral function:
\begin{align*}
I&=\int_0^1\frac{\left(1+x^2\right)\log(1+x)}{x^4-x^2+1}dx\\
&=\int_0^1\left(\frac{\omega}{1+\omega^2 x^2}+\frac{\bar{\omega}}{1+\bar{\omega}^2 x^2}\right)\log\left(1+x\right)dx\\
&=\left.\left(\arctan(\omega x)+\arctan(\bar{\omega}x)\right)\log(1+x)\right|_0^1-\int_0^1\frac{\arctan(\omega x)+\arctan(\bar{\omega} x)}{1+x}dx\\
&=\frac{\pi}{2}\log 2-\operatorname{Ti_2}(\omega,\omega)-\operatorname{Ti_2}(\bar{\omega},\bar{\omega})
\end{align*}
A result (3.27) from Polylogarithms and Associated Functions reads:
$$\operatorname{Ti_2}(a,a)+\operatorname{Ti_2}(a^{-1},a^{-1})=\frac{\pi}{2}\log\left(\frac{2}{\sqrt{1+a^2}}\right)+\arctan(a)\log(a)
$$
Thus letting $a=\omega$, we get:
$$I=\frac{\pi}{2}\log\left(\sqrt{1+\omega^2}\right)-\arctan(\omega)\log(\omega)=\frac{\pi}{6}\log\left(2+\sqrt{3}\right)
$$

This approach also produces a general result, for when $-\frac{\pi}{2}<\varphi<\frac{\pi}{2}$:
$$\int_0^1\frac{\left(1+x^2\right)\log(1+x)}{x^4+2\cos(2\varphi)x^2+1}dx=\frac{\pi}{8\cos\varphi}\log(2\cos\varphi)+\frac{\varphi}{4\cos\varphi}\log\tan\left(\frac{\varphi}{2}+\frac{\pi}{4}\right)
$$
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