Integrate $\Omega=\int_{-\infty}^{\infty}\frac{\operatorname{arccot}(x)}{x^4+x^2+1}dx$ A friend of mine got me the problem proposed by Vasile Mircea Popa from Romania, which was published in the Romanian mathematical Magazine. The problem is to find:

$$\Omega=\int_{-\infty}^{\infty}\frac{\operatorname{arccot}(x)}{x^4+x^2+1}dx$$

As per the Wolfram alpha the evaluated value is found to be $0$. The reason is since  $\operatorname{arccot}(x)=-\operatorname{arccot}(-x)$ for all $x\in\mathbb C^+$ is an odd function.
However, the next answer obtained is $\frac{\pi^2}{ 2\sqrt{3}}$ where  the relations  $\text{arccot}(x)=\frac{\pi}{2}-\operatorname{arctan}(x)\cdots(1)$ is used keeping in the view of principal branch of $\operatorname{arccot}(x)$. The works is as follows: $$\Omega=\int_{-\infty}^{\infty}\frac{\frac{\pi}{2}-\operatorname{arctan}(x)}{x^4+x^2+1}dx=\frac{\pi}{2}\int_{-\infty}^{\infty}\frac{dx}{x^4+x^2+1}-\underbrace{\int_{-\infty}^{\infty}\frac{\operatorname{arctan}(x)}{x^4+x^2+1}}_{\text{odd function}}dx\\\overbrace{=}^{xy=1}\frac{\pi}{2}\int_{-\infty}^{\infty}\frac{x^2 dx}{x^4+x^2+1}=\frac{\pi}{2}\int_{-\infty}^{\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^{2}+3}$$
then, by Cauchy Schlömilch transformation
(Special case of Glasser's Masters  theorem) we obtain $$\Omega= \frac{\pi}{2}\int_{-\infty}^{\infty}\frac{dx}{x^2+3}=\frac{\pi^2}{2\sqrt{3}}$$
Note that former integral can be solved without using aforementioned theorem, by the partial fraction of $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$.
My question is, Which of the above work is correct?
In my view the first work is correct. In  the second working,
Is the use of Maclaurin series done correctly?
 A: There are two main definitions of $\text{arccot}$ that people use.  The one employed by WolframAlpha is that $\text{arccot}$ is the inverse of $\cot:\left(-\dfrac{\pi}{2},+\dfrac{\pi}{2}\right]\to\mathbb{R}$, so that $\text{arccot}$ is an odd function on $\mathbb{R}_{\neq 0}$.  As in Botond's now deleted answer (which, I hope, would be undeleted),
$$\arctan(x)+\text{arccot}(x)=\dfrac{\pi}{2}\,\text{sign}(x)$$
for all $x\in\mathbb{R}$, where the sign function $\text{sign}:\mathbb{R}\to\{-1,+1\}$ uses the convention that $\text{sign}(0)=1$.  For a plot of this version of $\text{arccot}$, see here.
The other definition is that $\text{arccot}$ is the inverse of $\cot:(0,\pi)\to \mathbb{R}$, which makes $\text{arccot}$ satisfy
$$\arctan(x)+\text{arccot}(x)=\dfrac{\pi}{2}\text{ for all }x\in\mathbb{R}\,.$$
This is the definition I prefer because this version of $\text{arccot}$ is continuous and differentiable (see a plot here).  Furthermore, it aligns with other inverse trigonometric function identities:
$$\arcsin(x)+\arccos(x)=\dfrac{\pi}{2}\text{ for all }x\in[-1,+1]$$
and
$$\text{arcsec}(x)+\text{arccsc}(x)=\dfrac{\pi}{2}\text{ for all }x\in(-\infty,-1]\cup[+1,+\infty)\,.$$
