反三角函数（2）

\[\Large\displaystyle \int_0^{\infty} \frac{(1-x^2)\arctan x^2}{1+4x^2+x^4}\, {\rm d}x \]

Solution
What comes to mind is to maybe write the integrand as:

\[\Large\displaystyle\frac{\sqrt{3}-1}{2}\int_{0}^{\infty}\frac{\tan^{-1}(x^{2})}{x^{2}+2-\sqrt{3}}dx-\frac{\sqrt{3}+1}{2}\int_{0}^{\infty}\frac{\tan^{-1}(x^{2})}{x^{2}+2+\sqrt{3}}dx \]

Now, deal with

\[\Large\displaystyle \int_{0}^{\infty}\frac{\tan^{-1}(x^{2})}{x^{2}+b}dx\tag{1} \]

I think ben's idea of contours may be a cool way to go in this case.
Note the pole at \(\sqrt{b}i\) and \(\tan^{-1}(x^{2})\) has singularities at \(\pm \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\)

EDIT:
Please check my reasoning and feel free to add your part.
I looked at this and came up with something rough. I need to tweak it. The branch of arctan and all that gets me sideways. But, something looks to be coming together.
Please, feel free to finish the idea.

\[\Large\displaystyle f(z)=\frac{\tan^{-1}(z^{2})}{z^{2}+b} \]

Using the expansion at (1), the residues at the arctan singularity at \(\pm \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\) is 0.
With the residue at \(z=\sqrt{2-\sqrt{3}}i\) we then have:

\[\Large\displaystyle 2\pi i \cdot \frac{1}{2\sqrt{2-\sqrt{3}}i}\cdot \frac{\sqrt{3}-1}{2}\cdot \frac{5\pi}{12}=\frac{5\pi^{2}\sqrt{2}}{24} \]

with the residue at \(z=\sqrt{2+\sqrt{3}}i\), we have:

\[\Large\displaystyle 2\pi i \cdot \frac{1}{2\sqrt{2+\sqrt{3}}i}\cdot \frac{\sqrt{3}+1}{2}\cdot \frac{7\pi}{12}=\frac{7\pi^{2}\sqrt{2}}{24} \]

subtract per the expansion at the top and arrive at:

\[\Large\displaystyle \frac{5\pi^{2}\sqrt{2}}{24}-\frac{7\pi^{2}\sqrt{2}}{24}=\frac{-\pi^{2}\sqrt{2}}{12} \]

Now, it appears we have to divide by 2 to arrive at the final answer. I believe is due to the symmetry of the integrand:
That is, something like