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.
 A: 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
$$\int_{0}^{\infty} \frac{x^{2}}{(1+x^{4})^{2}} d x
= \frac14\int_{0}^{\infty}
\frac{x^{2}}{1+x^{4}} d x=\frac14 \cdot \frac\pi{2\sqrt2}=\frac\pi{8\sqrt2}
$$
A: You can use residue theorem. Start by extending the integral from $-\infty$ to $+\infty$, which you can do since the function is even.
Then, use the semi-circle in the complex plane for integration: $\Gamma(\theta) = Re^{i\theta}$, with $\theta  \in [0,\pi]$. You will get the desired result in the limit $R\rightarrow +\infty$.
In that limit, the integral in the semi-circle vanishes due to Jordan Lemma. The residues inside the contour are $e^{i\pi/4}$ and $e^{i3\pi/4}$, with Res$\big(\frac{z^2}{(z^4+1)^2},z=e^{i\pi/4}\big)=\frac{e^{3i\pi/4}}{16}$ and Res$\big(\frac{z^2}{(z^4+1)^2},z=e^{i3\pi/4}\big)=\frac{e^{5i\pi/4}}{16}$. The result is then 
$\int_{0}^{+\infty}dx \frac{x^2}{(x^4+1)^2} =\frac{1}{2} \int_{-\infty}^{+\infty}dx \frac{x^2}{(x^4+1)^2}=\pi i \sum \text{Res} = \frac{\pi i}{8}(e^{5i\pi/4} + e^{3i\pi/4} ) = \frac{\pi}{8\sqrt{2}} = \frac{\sqrt{2}\pi}{16} $
A: 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
