Integral in the form of an algebraic equation.

$\alpha =\frac{\int_0^{\infty } \frac{x^2+x+1}{\left(x^3+x+1\right) \sqrt{x}} \, dx}{\pi } \Longrightarrow \frac{1}{2} \sqrt{\frac{28}{93} \sqrt[3]{\frac{2}{47-3 \sqrt{93}}}+\frac{1}{93} 2^{2/3} 2 \sqrt[3]{47-3 \sqrt{93}}+24 \sqrt{\frac{6}{31 \left(-14 \sqrt[3]{\frac{2}{47-3 \sqrt{93}}}-2^{2/3} \sqrt[3]{47-3 \sqrt{93}}+40\right)}}+\frac{160}{93}}+\frac{1}{\sqrt{\frac{186}{-14 \sqrt[3]{\frac{2}{47-3 \sqrt{93}}}-2^{2/3} \sqrt[3]{47-3 \sqrt{93}}+40}}}$

$31 \alpha ^4-40 \alpha ^2-24 \alpha \Longrightarrow 4$

And, you should have seen the "herd of elephants" in the middle!

Addendum: $\text{FullSimplify}\left[\text{Root}\left[31 \text{\$\#\$1}^4-40 \text{\$\#\$1}^2-24 \text{\$\#\$1}-4\&,2\right]-\frac{\int_0^{\infty } \frac{x^2+x+1}{\left(x^3+x+1\right) \sqrt{x}} \, dx}{\pi }\right] \Longrightarrow 0$

The result above states that there is $0$ difference between the $2^{nd}$ root of $31 \alpha^4-40 \alpha^2-24 \alpha-4=0$ and the result of the integral.

Q.E.D. The answer is correct. How the author derived this problem, I have not idea.

I wrote up a solution in a a Math.SE posting , so let me simply outline the solution.

(Step 1) Write $p(x) = x^2 + x + 1$ and $q(x) = x^3 + x + 1$ . Choosing the branch cut of $\sqrt{\cdot}$ as $[0,\infty)$ and utilizing the keyhole contour, $\alpha = \sum_{x : q(x) = 0} \frac{ip(x)}{q'(x)\sqrt{x}}$ Since there are 3 distinct complex zeros of $q(x)$ , there are 3 summands, which we denote by $z_1$ , $z_2$ , and $z_3$ . Obviously, $\alpha = z_1 + z_2 + z_3$ .

(Step 2) The idea in this step is to identify the polynomial having zeros $z_1^2, z_2^2, z_3^2$ . Introduce a new variable $y$ by $y = \left( \frac{ip(x)}{q'(x)\sqrt{x}} \right)^2 = -\frac{p(x)^2}{xq'(x)^2}.$ This $y$ satisfies the equation $p(x)^2 + yxq'(x)^2 = 0$ . Let $\mathcal{Z}$ denotes the set of all possible values of $y$ given $q(x) = 0$ , or equivalently, $\mathcal{Z} = \{z_1^2, z_2^2, z_3^2 \}$ . Then $q(x)$ and $p(x)^2 + yxq'(x)^2$ have a common zero in $\mathbb{C}$ if $y \in \mathcal{Z}$ , and so, by the property of the resultant ,

• $\operatorname{res}\left( q(x), p(x)^2 + yxq'(x)^2 \right) = 0$ whenever $y \in \mathcal{Z}$ , where the coefficient ring is assumed as $\mathbb{C}[y]$ for computing the resultant.

On the other hand, by a direct computation, $\operatorname{res}\left( q(x), p(x)^2 + yxq'(x)^2 \right) = 961y^3 - 620y^2 + 131y - 9.$ Since this is a cubic polynomial in $y$ , it can have at most 3 distinct zeros in $\mathbb{C}$ , and so, the above observation tells that $961y^3 - 620y^2 + 131y - 9 = 31^2 (y - z_1^2)(y - z_2^2)(y - z_3^2).$ In particular, by the Vieta's formula , $z_1^2 + z_2^2 + z_3^2 = \frac{620}{961} = \frac{20}{31}, \qquad z_1^2 z_2^2 + z_2^2 z_3^2 + z_3^2 z_1^2 = \frac{131}{961}, \qquad z_1^2 z_2^2 z_3^2 = \frac{9}{961} = \left(\frac{3}{31}\right)^2.$ Moreover, although non-trivial, it can also be shown that $z_1 z_2 z_3$ is positive, and so, $z_1 z_2 z_3 = \frac{3}{31}$ .

(Step 3) Now we are almost done. Squaring $\alpha$ first, $\alpha^2 = \left( z_1^2 + z_2^2 + z_3^2 \right) + 2\left( z_1z_2 + z_2z_3 + z_3z_1 \right) = \frac{20}{31}+ 2\left( z_1z_2 + z_2z_3 + z_3z_1 \right)$ Subtracting $\frac{20}{31}$ from both sides and squaring again, $\left( \alpha^2 - \frac{20}{31} \right)^2 = 4 \left( z_1^2 z_2^2 + z_2^2 z_3^2 + z_3^2 z_1^2 \right) + 8 z_1 z_2 z_3 (z_1 + z_2 + z_3) = \frac{524}{961} + \frac{24}{31}\alpha.$ Simplifying, it follows that $31 \alpha^4 - 40\alpha^2 - 24\alpha = 4$ as required.