Is it hard to tackle the integral $\int_{0}^{\infty} \frac{x^{2}}{\left(1+x^{4}\right)^{2}} d x?$ Putting $ \displaystyle=4, \alpha=2, n=2 $ in my post,
$$\int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}}dx=\frac{\pi}{m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right)$$
we can conclude that
$$
\begin{aligned}
I &=\frac{\pi}{4(2-1) !} \csc \frac{3 \pi}{4}\left(1-\frac{3}{4}\right) \\
&=\frac{\sqrt{2} \pi}{16}
\end{aligned}
$$
Question:
Is there any other method?
Your suggestion and alternative are warmly welcome.

Integrate both sides of the equation $\left(\frac{x^3}{1+x^4}\right)’= \frac{4x^2}{(1+x^4)^2}-\frac{x^2}{1+x^4}$
to obtain
$$\int_{0}^{\infty} \frac{x^{2}}{(1+x^{4})^{2}} d x
= \frac14\int_{0}^{\infty}
\frac{x^{2}}{1+x^{4}} d x=\frac14 \cdot \frac\pi{2\sqrt2}=\frac\pi{8\sqrt2}
$$
