<s>[INST] 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.
 [/INST] Letting $x\mapsto \frac{1}{x} $ yields
$$
I=\int_{0}^{\infty} \frac{x^{4}}{\left(x^{4}+1\right)^{2}} d x
$$
Adding them together gives $$
\begin{aligned}
2 I &=\int_{0}^{\infty} \frac{x^{2}+x^{4}}{\left(x^{4}+1\right)^{2}} d x \\
&=\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}\right)^{2}} d x \\
&=\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^{2}+2\right]^{2}}
\end{aligned}
$$
Letting $x-\frac{1}{x}=\sqrt{2} \tan \theta$, we have
$$
I=\frac{\sqrt{2}}{8} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} \theta d \theta= \frac{\sqrt{2} \pi}{16}
$$
We can conclude that
</s>