Recently, I met a difficult integral in the post
$$ \int_0^{\infty} \frac{x^2 \tan ^{-1} x}{1+x^2+x^4} d x
= \frac{\pi^2}{8 \sqrt{3}}+\frac{\pi}{24} \ln \left(\frac{2-\sqrt{3}}{2+\sqrt{2}}\right)+\frac{2}{3} G,
$$
where $G$ is the Catalan’s constant.
Then I am curious what happens when the power $2$ in the numerator is replaced by other non-negative integers.
Starting with $n=0$,
$$
\begin{aligned}
I_0&=\int_0^{\infty} \frac{\tan ^{-1} x}{1+x^2+x^4} d x \\
&=\int_0^{\infty} \frac{\tan ^{-1} \frac{1}{x}}{1+\frac{1}{x^2}+\frac{1}{x^4}} d x \\
&=\int_0^{\infty} \frac{x^2 \tan ^{-1} \frac{1}{x}}{1+x^2+x^4} d x \\
&=\int_0^{\infty} \frac{x^2\left(\frac{\pi}{2}-\tan ^{-1} x\right)}{1+x^2+x^4} d x \\
&=\frac{\pi}{2} \int_0^{\infty} \frac{x^2}{1+x^2+x^4} d x-I_2 \\
&= \frac{\pi^2}{4 \sqrt{3}}-\left[\frac{\pi^2}{8 \sqrt{3}}+\frac{\pi}{24} \ln \left(\frac{2-\sqrt{3}}{2+\sqrt{3}}\right)+\frac{2}{3} G\right] \\
&=\frac{\pi^2}{8 \sqrt{3}}-\frac{\pi}{24} \ln \left(\frac{2-\sqrt{3}}{2+\sqrt{3}}\right)-\frac{2}{3} G
\end{aligned}
$$
When $n=1,$
$$I_1=\int_0^{\infty} \frac{x \tan ^{-1} x}{1+x^2+x^4} dx $$
Let’s consider the parametrised integral
$$
J(a)=\int_0^{\infty} \frac{x \tan ^{-1}(ax)}{1+x^2+x^4} d x
$$
Differentiating $J(a)$ w.r.t. $a$ yields
$$
\begin{aligned}
J^{\prime}(a) & =\int_0^{\infty} \frac{x^2}{\left(1+x^2+x^4\right)\left(1+a^2 x^2\right)} d x \\
& =\frac{\pi}{2 \sqrt{3}}\left(\frac{1+a^2-\sqrt{3} a}{a^4-a^2+1}\right)
\end{aligned}
$$
where the last answer using post with replacement of $a$ by $a^2$.
Integrating back gives $$
\begin{aligned}
\int_0^{\infty} \frac{x \tan ^{-1} x}{1+x^2+x^4} d x & =J(1)-J(0) \\
& =\frac{\pi}{2 \sqrt{3}} \int_0^1 \frac{1+a^2-\sqrt{3} a}{a^4-a^2+1} d a\\&=\frac{\pi^2}{12 \sqrt{3}}
\end{aligned}
$$
When $n=3 $, I had tried various methods but failed, therefore I ask for any method to evaluate or prove the divergence of
$$
I_3=\int_0^{\infty} \frac{x^3 \tan ^{-1} x}{1+x^2+x^4} d x
$$
OR $$
I_n=\int_0^{\infty} \frac{x^n \tan ^{-1} x}{1+x^2+x^4} d x
$$
Your comments and solutions are highly appreciated.
Best Answer
Using Feynman's trick, $I_3$ cannot converge. $$J_3(a)=\int_0^{\infty} \frac{x^3 \tan ^{-1}(ax)}{1+x^2+x^4} \,dx$$ $$J_3'(a)=\int_0^{\infty}\frac{x^4}{\left(x^4+x^2+1\right) \left(a^2 x^2+1\right)}\,dx$$ Using partial fraction decomposition, the integrand is $$\frac{a^2 (-x)-1}{2 \left(a^4-a^2+1\right) \left(x^2+x+1\right)}+\frac{a^2 x-1}{2 \left(a^4-a^2+1\right) \left(x^2-x+1\right)}+\frac{1}{\left(a^4-a^2+1\right) \left(a^2 x^2+1\right)}$$ making $$J_3'(a)=\frac \pi 6 \frac{\sqrt{3}\, a^3-2 \sqrt{3}\, a+3}{\color{red}{a}\,\left(a^4-a^2+1\right)}$$ which cannot be integrated between $0$ and $1$ because of the $\color{red}{a}$.
Edit
There is no problem with $n=\frac 5 2$ $$J'_{5/2}(a)=\int_0^{\infty}\frac{x^{7/2}}{\left(x^4+x^2+1\right) \left(a^2x^2+1\right)}\,dx$$ $$J'_{5/2}(a)=\frac \pi 6 \frac {\left(3-\sqrt{3}\right) a^{5/2}-\left(3+\sqrt{3}\right) \sqrt{a}+3 \sqrt{2} } {(a^4-a^2+1)\sqrt a }$$ $$I_{5/2}=\frac{\pi }{4 \sqrt{3}} \left(\pi +4 \coth ^{-1}\left(\frac{4+\sqrt{2}}{\sqrt{6}}\right)\right)$$