Integrating $\int_{-\infty}^{\infty} \big(\frac{x^2}{x^2 + a}\big)^2 e^{-x^2/2} \, \mathrm{d}x$ I was wondering if it is possible to compute this integral in closed form:
$$
\int_{-\infty}^{\infty} \big(\frac{x^2}{x^2 + a}\big)^2 e^{-x^2/2} \, \mathrm{d}x
$$
I tried making a substitution with $s = x^2$ and also tried evaluating it in Wolfram Alpha, but was not successful.

Under substitution $\frac{x}{\sqrt{2}} \to x$ we get that your integral is equal to $I =2 \sqrt{2} \int_0^{\infty} \frac{x^4}{(x^2 + \xi)^2}e^{-x^2}\mathrm{d}x$ where $\xi = \frac{a}{2}$. Additionally, since $ \frac{x^4}{(x^2 + \xi)^2} = 1 - \frac{2\xi}{x^2 + \xi} + \frac{\xi^2}{\left(x^2 + \xi \right)^2}$, the problem reduces to solving
$$
I_n = \int_0^{\infty} \frac{e^{-x^2}}{(x^2 + \xi)^n}\, \mathrm{d}x,\ \ \quad \text{for}\quad   n=1,2
$$

For the first integral we get
\begin{align*}
I_1 & =\int_0^{\infty} \frac{e^{-x^2}}{x^2 + \xi}\, \mathrm{d}x\\ & = \int_0^{\infty}\int_0^{\infty} e^{-x^2}e^{-y\left(x^2+ \xi\right) }  \, \mathrm{d}y \, \mathrm{d}x \\
& = \int_0^{\infty}\int_0^{\infty} e^{-x^2(y+1)}e^{-y \xi }  \, \mathrm{d}x \, \mathrm{d}y 
\end{align*}
Introducing the change of coordinate system $x = \sqrt{\xi}\frac{u}{v}$ and $ y = \frac{v^2}{\xi} -1$ gives $v^2 -\xi= \xi y$ and $u^2 = x^2 (y+1)$ and thus
\begin{align*}
I_1 &= \int_{\sqrt{\xi}}^{\infty} \int_{0}^{\infty} e^{-u^2}e^{-v^2 +\xi} \begin{vmatrix} \sqrt{\xi}\frac{1}{v} & -\sqrt{\xi}\frac{u}{v^2} \\ 0 & \frac{2v}{\xi}\end{vmatrix}\, \mathrm{d}u \, \mathrm{d}v\\
& = \frac{2}{\sqrt{\xi}} e^{\xi} \int_{0}^{\infty}e^{-u^2}\mathrm{d}u \int_{\sqrt{\xi}}^{\infty}e^{-v^2}\mathrm{d}v\\
& = \frac{\pi}{2\sqrt{\xi}} e^{\xi} \mathrm{erfc}\left(\sqrt{\xi}\right)
\end{align*}
recalling that $\int_0^{\infty} e^{-t^2}\mathrm{d}t = \frac{\sqrt{\pi}}{2}$ and that $\int_z^{\infty} e^{-t^2}\mathrm{d}t = \frac{\sqrt{\pi}}{2}\mathrm{erfc}(z)$ by the definition of the complementary error function. Thus

$$\int_0^{\infty} \frac{e^{-x^2}}{x^2 + \xi}\, \mathrm{d}x =\frac{\pi}{2\sqrt{\xi}} e^{\xi} \mathrm{erfc}\left(\sqrt{\xi}\right) \qquad \text{for} \quad \xi >0  $$

For the second integral we'll use Feynman's trick. Since $\frac{\partial}{\partial \xi} \frac{e^{-x^2}}{x^2 + \xi} = - \frac{e^{-x^2}}{\left(x^2 + \xi\right)^2}$ we see that differentiating the previous result we can obtain $I_2$:
$$
-I_2 = \int_0^{\infty}\frac{\partial}{\partial \xi}  \frac{e^{-x^2}}{x^2 + \xi}\, \mathrm{d}x = \frac{\mathrm{d}}{\mathrm{d}\xi}\int_0^{\infty} \frac{e^{-x^2}}{x^2 + \xi}\, \mathrm{d}x =  \frac{\mathrm{d}}{\mathrm{d}\xi}\frac{\pi}{2\sqrt{\xi}} e^{\xi} \mathrm{erfc}\left(\sqrt{\xi}\right)
$$
By the F.T.C. we know that $\frac{\mathrm{d}}{\mathrm{d}z}\mathrm{erfc}(z) = - \frac{2}{\sqrt{\pi}}e^{-z^2}$ using the definition of the complementary error function. Taking the derivative we get that $\frac{\mathrm{d}}{\mathrm{d}\xi} \frac{e^\xi}{\sqrt{\xi}}\mathrm{erfc}\left(\sqrt{\xi}\right) = \frac{e^\xi (2\xi-1)}{2\xi^{\frac32}}\mathrm{erfc}\left(\sqrt{\xi}\right) - \frac{1}{\sqrt{\pi}\xi}  $ which allows us to conclude

$$\int_0^{\infty} \frac{e^{-x^2}}{\left(x^2 + \xi\right)^2}\, \mathrm{d}x =\frac{\sqrt{\pi}}{2\xi}  - \frac{\pi e^\xi (2\xi-1)}{4\xi^{\frac32}}\mathrm{erfc}\left(\sqrt{\xi}\right) \qquad \text{for} \quad \xi >0  $$


Finally, combining all of the above we can evaluate the original integral to be
\begin{align*}
I & = 2 \sqrt{2} \int_0^{\infty} \frac{x^4}{(x^2 + \xi)^2}e^{-x^2}\mathrm{d}x \\
& = 2 \sqrt{2} \left[\int_0^{\infty} e^{-x^2}\mathrm{d}x  - 2\xi \int_0^{\infty} \frac{e^{-x^2}}{x^2 + \xi} \mathrm{d}x + \xi^2\int_0^{\infty} \frac{e^{-x^2}}{\left(x^2 + \xi \right)^2}\mathrm{d}x  \right]\\
& = 2 \sqrt{2} \left[\frac{\sqrt{\pi}}{2} - 2\xi \frac{\pi}{2\sqrt{\xi}} e^{\xi} \mathrm{erfc}\left(\sqrt{\xi}\right)  + \xi^2\left( \frac{\sqrt{\pi}}{2\xi}  - \frac{\pi e^\xi (2\xi-1)}{4\xi^{\frac32}}\mathrm{erfc}\left(\sqrt{\xi}\right)\right) \right]\\
& =  (\xi+1)\sqrt{2\pi} - (2\xi+3) \pi e^\xi \sqrt{\frac{\xi}{2}}\mathrm{erfc}\left(\sqrt{\xi}\right)
\end{align*}
and recalling that $ \xi = \frac{a}{2}$ we get the final result
$$
\boxed{\int_{-\infty}^{\infty} \left(\frac{x^2}{x^2+a} \right)^2 e^{-\frac{x^2}{2}} \, \mathrm{d}x =\frac{1}{2}\left[(a+2)\sqrt{2\pi} - \sqrt{a}(a+3) \pi e^\frac{a}{2} \mathrm{erfc}\left(\sqrt{\frac{a}{2}}\right) \right] \qquad \text{for} \quad a>0}
$$
verifying WA's result posted in the comments.
