How to evaluate $\int_{0}^{\infty }\frac{x^{4}e^{-2x^{2}}}{\left ( 1+x^{2} \right )^{4}}\mathrm{d}x$ How to evaluate the following integral
$$\int_{0}^{\infty }\frac{x^{4}e^{-2x^{2}}}{\left ( 1+x^{2} \right )^{4}}\mathrm{d}x$$
Can we solve it without using complex analysis method?

We will illustrate a methodology for evaluating the integral of interest that is based on "Feynmann's Method" for differentiating under the integral.  
To begin, we augment the integral of interest by introducing the parameter $a$ in the exponential argument of the integrand.  let $I(a)$ denote the integral 
$$I(a)=\int_0^\infty \frac{x^4e^{-ax^2}}{(1+x^2)^4} \,dx\tag 1$$
Then, the integral of interest is simply given by
$$I(2)=\int_0^\infty \frac{x^4e^{-ax^2}}{(1+x^2)^4} \,dx$$
Using partial fraction expansion on the term $\frac{x^4e^{-ax^2}}{(1+x^2)^4}$ reveals
$$\frac{x^4}{(1+x^2)^4}=\frac{1}{(1+x^2)^2}-\frac{2}{(1+x^2)^3}+\frac{1}{(1+x^2)^4} \tag 2$$
whereby we can write $(1)$ as
$$\begin{align}
I(a)&=\int_0^\infty \frac{e^{-ax^2}}{(1+x^2)^2}\,dx-2\int_0^\infty \frac{e^{-ax^2}}{(1+x^2)^3}\,dx+\int_0^\infty \frac{e^{-ax^2}}{(1+x^2)^4}\,dx \\\\
&=e^a\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^2}\,dx-2e^a\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^3}\,dx+e^a\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^4}\,dx \tag 3
\end{align}$$
We can evaluate each of the $3$ integrals on the right-hand side of $(3)$ by differentiating under the integral.  In particular, we will evaluate the first integral on the right-hand side of $(3)$ and leave the evaluation of the second and third integrals as an exercise.
Let $F_n(a)$ denote the integral
$$F_n(a)=\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^n}\,dx$$
The integral $F_0(a)$ is Gaussian and easily evaluated; we find that $F_0(a)$ is 
$$\begin{align}
F_0(a)&=\int_0^\infty e^{-a(1+x^2)}\,dx\\\\
&=\frac{\sqrt \pi}{2}\frac{e^{-a}}{\sqrt a}\end{align}$$

INTEGRATING $F_0(a)$
Noting $F_1'(a)=-F_0(a)$, with $F_1(0)=\pi/2$, yields
$$\begin{align}
F_1(a)&=\int_0^\infty \frac{e^{-a(1+x^2)}}{1+x^2}\,dx\\\\
&=\frac\pi 2-\int_0^a F_0(y)\,dy\\\\
&=\frac{\pi }{2}-\frac{\sqrt \pi}{2}\int_0^a \frac{e^{-y}}{\sqrt y}\,dy\\\\
&=\frac{\pi}{ 2}-\frac{\pi}{2}\frac{2}{\sqrt \pi}\int_0^a e^{-t^2}\,dt\\\\
&=\frac{\pi}{2}\text{erfc}(\sqrt a)
\end{align}$$

INTEGRATING $F_1(a)$
Similarly, since $F_2'(a)=-F_1(a)$, with $F_2(0)=\pi/4$, we have
$$\begin{align}
F_2(a)&=\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^2}\,dx\\\\
&=\frac\pi 4-\int_0^a F_1(x)\,dx\\\\
&=\frac\pi 4-\frac{\pi}{2}\int_0^a \text{erfc}(\sqrt x)\,dx\\\\
&=\frac\pi 4-\pi \int_0^{\sqrt a} x\,\text{erfc}(x)\,dx\\\\
&=\frac\pi 4-\pi\left(\frac a2\text{erfc}(\sqrt a)-\frac12 \int_0^{\sqrt{a}} x^2\,\text{erfc}'(x)\,dx\right)\\\\
&=\frac\pi 4-\frac\pi 2 a\,\text{erfc}(\sqrt a)-\frac\pi 2 \frac{2}{\sqrt \pi}\int_0^\sqrt a x^2e^{-x^2}\,dx\\\\
&=\frac\pi 4-\frac\pi 2 a\,\text{erfc}(\sqrt a)-\frac\pi 2 \frac{2}{\sqrt \pi}\left(\frac{\sqrt \pi}{4}\text{erf}(\sqrt a)-\frac {\sqrt a}2  e^{-a}\right)\\\\
&=\frac{\pi}{4}\left(1-2a\right)\text{efrc}(\sqrt a)+\frac{\sqrt \pi}{2}\sqrt a e^{-a}
\end{align}$$
Thus, the first term on the right-hand side of $(3)$ is $e^2F_2(2)$.
The evaluation of $F_3(2)$ and $F_4(2)$ follow along the same methodology and is left as an exercise.
