Finding $\int^1_0 \frac{x\,\mathrm dx}{((2x-1)\sqrt{x^2+x+1} + (2x+1)\sqrt{x^2-x+1})\sqrt{x^4+x^2+1}}$ $$\int^1_0 \frac{x\,\mathrm dx}{((2x-1)\sqrt{x^2+x+1} + (2x+1)\sqrt{x^2-x+1})\sqrt{x^4+x^2+1}}$$
I've tried :

*

*Substitution:

*

*$t = x^2$

*$x = \cos{t}$



*Definite Integral Properties like $a+b-x$ (which messes up the integral symmetry)

*Partial Integration by breaking integral into into $\frac{x}{\sqrt{\left(x^4+x^2+1\right)}}$

*Partial Fractions which are very ugly

As per wolframAlpha, a beautiful elementary closed form exists
$$
\frac{\sqrt{\left(x^4+x^2+1\right)}}{3}\left(\frac{1}{\sqrt{\left(x^2-x+1\right)}}- \frac{1}{\sqrt{\left(x^2+x+1\right)}}\right)
$$
I'll be happy with solution of just definite integral, bonus points if you get closed form solution too.
 A: 
Let $\mathcal{I}$ denote the value of the following definite integral:
$$\mathcal{I}:=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left[\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}\right]\sqrt{x^{4}+x^{2}+1}}\approx0.2440169.$$
Observe that $\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)=x^{4}+x^{2}+1$ and
$\left[\left(2x+1\right)\sqrt{x^{2}-x+1}-\left(2x-1\right)\sqrt{x^{2}+x+1}\right]\left[\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}\right]=6x$.
Then,
$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left[\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}\right]\sqrt{x^{4}+x^{2}+1}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}}\\
&~~~~~\times\frac{1}{\sqrt{\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\left(2x+1\right)\sqrt{x^{2}-x+1}-\left(2x-1\right)\sqrt{x^{2}+x+1}}{6}\\
&~~~~~\times\frac{1}{\sqrt{x^{2}-x+1}\sqrt{x^{2}+x+1}}\\
&=\frac13\int_{0}^{1}\mathrm{d}x\,\frac{\left(2x+1\right)\sqrt{x^{2}-x+1}-\left(2x-1\right)\sqrt{x^{2}+x+1}}{2\sqrt{x^{2}-x+1}\sqrt{x^{2}+x+1}}\\
&=\frac13\int_{0}^{1}\mathrm{d}x\,\left[\frac{\left(2x+1\right)}{2\sqrt{x^{2}+x+1}}-\frac{\left(2x-1\right)}{2\sqrt{x^{2}-x+1}}\right]\\
&=\frac13\int_{0}^{1}\mathrm{d}x\,\frac{d}{dx}\left[\sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1}\right]\\
&=\frac{\sqrt{3}-1}{3}.\\
\end{align}$$

