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

Question

$$\int^1_0 \frac{x\,\mathrm dx}{((2x-1)\sqrt{x^2+x+1} + (2x+1)\sqrt{x^2-x+1})\sqrt{x^4+x^2+1}}$$

I've tried :

1. Substitution:
   • $t = x^2$
   • $x = \cos{t}$
2. Definite Integral Properties like $a+b-x$ (which messes up the integral symmetry)
3. Partial Integration by breaking integral into into $\frac{x}{\sqrt{\left(x^4+x^2+1\right)}}$
4. 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.

Answer 1

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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,

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left[\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}\right]\sqrt{x^{4}+x^{2}+1}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}}\\ &~~~~~\times\frac{1}{\sqrt{\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\left(2x+1\right)\sqrt{x^{2}-x+1}-\left(2x-1\right)\sqrt{x^{2}+x+1}}{6}\\ &~~~~~\times\frac{1}{\sqrt{x^{2}-x+1}\sqrt{x^{2}+x+1}}\\ &=\frac13\int_{0}^{1}\mathrm{d}x\,\frac{\left(2x+1\right)\sqrt{x^{2}-x+1}-\left(2x-1\right)\sqrt{x^{2}+x+1}}{2\sqrt{x^{2}-x+1}\sqrt{x^{2}+x+1}}\\ &=\frac13\int_{0}^{1}\mathrm{d}x\,\left[\frac{\left(2x+1\right)}{2\sqrt{x^{2}+x+1}}-\frac{\left(2x-1\right)}{2\sqrt{x^{2}-x+1}}\right]\\ &=\frac13\int_{0}^{1}\mathrm{d}x\,\frac{d}{dx}\left[\sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1}\right]\\ &=\frac{\sqrt{3}-1}{3}.\\ \end{align}$$

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