How to evaluate $\int_{0}^{\infty }\frac{x^{4}e^{-2x^{2}}}{\left ( 1+x^{2} \right )^{4}}\mathrm{d}x$

I would use the identity $$(1+X)^{-\gamma}=\frac{1}{\Gamma(\gamma)}\int_0^\infty d\xi\ \xi^{\gamma-1}e^{-\xi}e^{-\xi X}$$ to write $$ \int_{0}^{\infty }\frac{x^{4}e^{-2x^{2}}}{\left ( 1+x^{2} \right )^{4}}\mathrm{d}x =\frac{1}{\Gamma(4)}\int_{0}^\infty d\xi\ \xi^3 e^{-\xi} \int_0^\infty dx\ x^4 e^{-2 x^2-\xi x^2}=\frac{1}{\Gamma(4)}\int_0^\infty d\xi \xi^3 e^{-\xi}\frac{3 \sqrt{\pi }}{8 (\xi +2)^{5/2}}\ , $$ which seems much nicer. Change variables to $\xi+2=t$ to get $$ \frac{3\sqrt{\pi}e^{2}}{8\Gamma(4)}\int_2^\infty dt (t-2)^3 \frac{e^{-t}}{t^{5/2}}\ . $$ Expanding $(t-2)^3$, you are left with four elementary integrals.

Define $$ f_k(a)=\int_0^\infty\frac{e^{-ax^2}\,\mathrm{d}x}{\left(1+x^2\right)^k}\tag{1} $$ Then we have $$ \int_0^\infty\frac{x^4e^{-2x^2}}{(1+x^2)^4}\,\mathrm{d}x=f_4''(2)\tag{2} $$ From $(1)$, we get the recursion $f_k(a)-f_k'(a)=f_{k-1}(a)$, which means $$ f_k(a)=e^a\int_a^\infty e^{-t}f_{k-1}(t)\,\mathrm{d}t\tag{3} $$ Starting with $$ f_0(a)=\frac12\sqrt{\frac\pi{a}}\tag{4} $$ and applying $(3)$ $4$ times, we get $$ f_4(a)=\frac{\sqrt{\pi a}}{48}\left(4a^2-8a+15\right)-\frac{\pi e^a}{96}\left(8a^3-12a^2+18a-15\right)\operatorname{erfc}\left(\sqrt{a}\right)\tag{5} $$ Applying $(2)$ to $(5)$, we get $$ \bbox[5px,border:2px solid #C0A000]{\int_0^\infty\frac{x^4e^{-2x^2}}{(1+x^2)^4}\,\mathrm{d}x =\frac{17}{16}\sqrt{2\pi}-\frac{241}{96}\pi e^2\operatorname{erfc}\left(\sqrt2\right)}\tag{6} $$ where $$ \operatorname{erfc}(a)=\frac2{\sqrt\pi}\int_a^\infty e^{-x^2}\,\mathrm{d}x\tag{7} $$

Lets apply Plancherel's theorem

$$ \int_{\mathbb{R}} f(x)g(x)dx= \int_{\mathbb{R}}\hat{f}(k)\hat{g}(k)dk \quad (*) $$

Here the hat stands for the Fourier transform of the corresponding quantity Note that this is possible because we are dealing with an even function and we can drop some compelx conjugations because everything is real.

Let's set $f(x)=e^{-2x^2}$ and $g(x)=\frac{x^4}{(1+x^2)^4}$. Then we have (both FT are quiet standard,complete the square for the first and use residue theorem for the second)

$$ \hat{f}(k)=\frac{1}{2}e^{-k^2/8}\\ \hat{g}(k)=\frac{\sqrt{\pi}}{48\sqrt{2}}\left(\Theta(k)e^{-k}(k^3-6 k^2+3k+3)-\Theta(-k)e^{k}(k^3+6 k^2+3k-3)\right) $$

where $\Theta(k)$ denotes the Heaviside function. I don't have the time a the moment to finish the rather tedious but straightforward calculation of (*) from here on but i'm sure u can bring it home if u keep in mind the definition of the error function