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.
Integral $\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$
closed-formdefinite integralsintegration
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Best Answer
\begin{align}J&=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx\\ &=\int_0^1\int_0^1 \frac{x(x^2+1)}{(x^4-x^2+1)(1+xt)}\,dt\,dx\\ &=-\int_0^1\int_0^1 \frac{t(t^2+1)}{(t^4-t^2+1)(1+xt)}\,dt\,dx+\int_0^1\int_0^1 \frac{t^3+t^2x-\sqrt{3}t^2-\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2-\sqrt{3}x+1)}\,dt\,dx+\\ &\int_0^1\int_0^1 \frac{t^3+t^2x+\sqrt{3}t^2+\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2+\sqrt{3}x+1)}\,dt\,dx\\ &=-J+\int_0^1\int_0^1 \frac{t^3+t^2x-\sqrt{3}t^2-\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2-\sqrt{3}x+1)}\,dt\,dx+\\ &\int_0^1\int_0^1 \frac{t^3+t^2x+\sqrt{3}t^2+\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2+\sqrt{3}x+1)}\,dt\,dx\\ \end{align}
Since,
\begin{align}A_1&=\int_0^1 \frac{t}{t^4-t^2+1}\,dt\\ &=\frac{1}{\sqrt{3}}\left[\arctan\left(\frac{2t^2-1}{\sqrt{3}}\right)\right]_0^1\\ &=\frac{\pi}{3\sqrt{3}}\\ A_3&=\int_0^1 \frac{t^3}{t^4-t^2+1}\,dt\\ &=\frac{1}{12}\left[2\sqrt{3}\arctan\left(\frac{2t^2-1}{\sqrt{3}}\right)+3\ln(t^4-t^2+1)\right]_0^1\\ &=\frac{\pi}{6\sqrt{3}}\\ B_1&=\int_0^1 \frac{1}{x^2-\sqrt{3}x+1}\,dx\\ &=2\Big[\arctan\left(2x-\sqrt{3}\right)\Big]_0^1\\ &=\frac{5\pi}{6}\\ B_2&=\int_0^1 \frac{1}{x^2+\sqrt{3}x+1}\,dx\\ &=2\Big[\arctan\left(2x+\sqrt{3}\right)\Big]_0^1\\ &=\frac{\pi}{6}\\ C_1&=\int_0^1 \frac{x}{x^2-\sqrt{3}x+1}\,dx\\ &=\Big[\frac{1}{2}\ln\left(x^2-\sqrt{3}x+1\right)+\sqrt{3}\arctan\left(2x-\sqrt{3}\right)\Big]_0^1\\ &=\frac{1}{2}\ln\left(2-\sqrt{3}\right)+\frac{5}{4\sqrt{3}}\pi\\ C_2&=\int_0^1 \frac{x}{x^2+\sqrt{3}x+1}\,dx\\ &=\Big[\frac{1}{2}\ln\left(x^2+\sqrt{3}x+1\right)-\sqrt{3}\arctan\left(2x+\sqrt{3}\right)\Big]_0^1\\ &=\frac{1}{2}\ln\left(2+\sqrt{3}\right)-\frac{1}{4\sqrt{3}}\pi\\ A_2&=\int_0^1 \frac{t^2}{t^4-t^2+1}\,dt\\ &=\int_0^1 \frac{t^2+1}{t^4-t^2+1}\,dt-\int_0^1 \frac{1}{t^4-t^2+1}\,dt\\ &=\left[\arctan\left(\frac{x}{1-x^2}\right)\right]_0^1-\frac{1}{2\sqrt{3}}\int_0^1 \frac{x}{x^2+\sqrt{3}x+1}\,dx+\frac{1}{2\sqrt{3}}\int_0^1 \frac{x}{x^2-\sqrt{3}x+1}\,dx-\\ &\frac{1}{2}\int_0^1 \frac{1}{x^2+\sqrt{3}x+1}\,dx-\frac{1}{2}\int_0^1 \frac{1}{x^2-\sqrt{3}x+1}\,dx\\ &=\frac{\pi}{2}-\frac{1}{2\sqrt{3}}C_2+\frac{1}{2\sqrt{3}}C_1-\frac{1}{2}B_2-\frac{1}{2}B_1\\ &=\frac{\pi}{4}-\frac{1}{2\sqrt{3}}\ln\left(2+\sqrt{3}\right)\\ A_0&=\int_0^1 \frac{1}{t^4-t^2+1}\,dt\\ &=\int_0^1 \frac{1+t^2}{t^4-t^2+1}\,dt-\int_0^1 \frac{t^2}{t^4-t^2+1}\,dt\\ &=\left[\arctan\left(\frac{x}{1-x^2}\right)\right]_0^1-A_2\\ &=\frac{\pi}{4}+\frac{1}{2\sqrt{3}}\ln\left(2+\sqrt{3}\right)\\ \end{align}
then,
\begin{align} 2J&=\left(\frac{1}{2}A_3B_1+\frac{1}{2}A_2C_1-\frac{\sqrt{3}}{2}A_2B_1-\frac{\sqrt{3}}{2}A_1C_1+\frac{1}{2}A_1B_1+\frac{1}{2}A_0C_1\right)+\\ &\left(\frac{1}{2}A_3B_2+\frac{1}{2}A_2C_2+\frac{\sqrt{3}}{2}A_2B_2+\frac{\sqrt{3}}{2}A_1C_2+\frac{1}{2}A_1B_2+\frac{1}{2}A_0C_2\right)\\ \end{align}
Since,
\begin{align}\frac{1}{2}A_3B_1&=\frac{5\pi^2}{72\sqrt{3}}\\ \frac{1}{2}A_2C_1&=\frac{5\pi^2}{32\sqrt{3}}+\frac{\pi}{16}\ln\left(2-\sqrt{3}\right)-\frac{1}{8\sqrt{3}}\ln\left(2-\sqrt{3}\right)\ln\left(2+\sqrt{3}\right)-\frac{5\pi}{48}\ln\left(2+\sqrt{3}\right)\\ -\frac{\sqrt{3}}{2}A_2B_1&=\frac{5\pi}{24}\ln\left(2+\sqrt{3}\right)-\frac{5\pi^2}{16\sqrt{3}}\\ -\frac{\sqrt{3}}{2}A_1C_1&=-\frac{\pi}{12}\ln\left(2-\sqrt{3}\right)-\frac{5\pi^2}{24\sqrt{3}}\\ \frac{1}{2}A_1B_1&=\frac{5\pi^2}{36\sqrt{3}}\\ \frac{1}{2}A_0C_1&=\frac{\pi}{16}\ln\left(2-\sqrt{3}\right)+\frac{5\pi^2}{32\sqrt{3}}+\frac{1}{8\sqrt{3}}\ln\left(2-\sqrt{3}\right)\ln\left(2+\sqrt{3}\right)+\frac{5\pi}{48}\ln\left(2+\sqrt{3}\right)\\ \frac{1}{2}A_3B_2&=\frac{\pi^2}{72\sqrt{3}}\\ \frac{1}{2}A_2C_2&=\frac{\pi}{12}\ln\left(2+\sqrt{3}\right)-\frac{\pi^2}{32\sqrt{3}}-\frac{1}{8\sqrt{3}}\ln^2\left(2+\sqrt{3}\right)\\ \frac{\sqrt{3}}{2}A_2B_2&=\frac{\pi^2}{16\sqrt{3}}-\frac{\pi}{24}\ln\left(2+\sqrt{3}\right)\\ \frac{\sqrt{3}}{2}A_1C_2&=\frac{\pi}{12}\ln\left(2+\sqrt{3}\right)-\frac{\pi^2}{24\sqrt{3}}\\ \frac{1}{2}A_1B_2&=\frac{\pi^2}{36\sqrt{3}}\\ \frac{1}{2}A_0C_2&=\frac{1}{8\sqrt{3}}\ln^2\left(2+\sqrt{3}\right)+\frac{\pi}{24}\ln\left(2+\sqrt{3}\right)-\frac{\pi^2}{32\sqrt{3}}\\ \end{align}
Therefore,
\begin{align}2J&=\frac{\pi}{24}\ln\left(2-\sqrt{3}\right)+\frac{3\pi}{8}\ln\left(2+\sqrt{3}\right)\end{align}
Since,
\begin{align}\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\end{align}
Therefore,
\begin{align}2J&=-\frac{\pi}{24}\ln\left(2+\sqrt{3}\right)+\frac{3\pi}{8}\ln\left(2+\sqrt{3}\right)\\ &=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right) \end{align}
Thus,
\begin{align}\boxed{J=\frac{\pi}{6}\ln\left(2+\sqrt{3}\right)} \end{align}