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?
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 $$\text{arcsec}(x)+\text{arccsc}(x)=\dfrac{\pi}{2}\text{ for all }x\in(-\infty,-1]\cup[+1,+\infty)\,.$$