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.

This is too long for a comment.
Since, in comments, you already received the answer and a good hint, let me address the more general case of
$$I_n= \int_{-\infty}^{+\infty} \left(\frac{x^2}{x^2+a}\right)^n\, e^{-\frac{x^2}{2}}\,dx=2\int_{0}^{+\infty} \left(\frac{x^2}{x^2+a}\right)^k\, e^{-\frac{x^2}{2}}\,dx$$
The idea is to write
$$\frac{x^2}{x^2+a}=\frac{x^2+a-a}{x^2+a}=1-\frac {a}{x^2+a}$$ and to use the binomial expansion to face known integrals since using Tricomi's confluent hypergeometric function
$$\int_0^\infty  \left(a+x^2\right)^{-n}\, e^{-\frac{x^2}{2}}\,dx=\frac 1 {2 ^n}\sqrt{\frac \pi 2 } \, U\left(n,n+\frac{1}{2},\frac{a}{2}\right)$$
They write
$$I_n=\frac{\sqrt{2\pi }}  {2^n\,n!}\Bigg[ P_{n}- {  \sqrt{\frac{a\pi}2}\,e^{a/2} } \, \text{erfc}\left(\sqrt{\frac{a}{2}}\right)\,Q_{n}\Bigg]$$
where the first polynomials in $a$ are
$$\left(
\begin{array}{ccc}
 n & P_n & Q_n \\
 0 & 1 & 0 \\
 1 & 2 & 2 \\
 2 & 4 a+8 & 4 a+12 \\
 3 & 6 a^2+54 a+48 & 6 a^2+60 a+90 \\
 4 & 8 a^3+160 a^2+696 a+384 & 8 a^3+168 a^2+840 a+840
\end{array}
\right)$$
