略解ですみません
$$I=\int_{0}^{1} \frac{\arctan({x})}{1+x^2} \left(\frac{x^4-6x^2+1}{(1+x^2)^2}\right)^2 dx$$
$x=tan (\theta)$
$$=\int_{0}^{\pi/4}\theta \left( \frac{\tan^4({\theta})-6\tan^2({\theta})+1}{(1+\tan^2({\theta}))^2} \right)^2 d\theta$$
$4$倍角の公式
$$=\int_{0}^{\pi/4}\theta \cos^2({4\theta} )d\theta$$
$King's Property$
$$=\int_{0}^{\pi/4} \left( \frac{\pi}{4}-\theta \right)\cos^2({4\theta}) d\theta$$
$\Rightarrow$
$$2I=\frac{\pi}{4} \int_{0}^{\pi/4}\cos^2({4\theta}) d\theta$$
$$=\frac{\pi}{4}\cdot\frac{\pi}{8}$$