Fourier Transform of complicated product: $(1+x)^2 e^{-x^2/2}$

If I understand the comments correctly, the remaining problem seems to be to find the Fourier transform of $$\color{blue}{f(x) = e^{-x^2/2}}.$$ You can do this by completing the squares and substitution, as outlined in the comments, but I prefer this argument:

Let's compute formally first, deferring the justification of the interchange of differentiation and integration in $(!!)$ to the end of the answer: $$ \begin{align*} \frac{d}{dk} \widehat{f}(k) &= \frac{d}{dk} \int_{-\infty}^{\infty} e^{-x^2/2}e^{-ikx}\,dx \\ &\!\color{red}{\stackrel{(!!)}{=}} \int_{-\infty}^\infty e^{-x^2/2}\, (-ix)\, e^{-ikx}\,dx \tag{!!} \\ &= \int_{-\infty}^\infty i \left( \frac{d}{dx}e^{-x^2/2} \right) \, e^{-ikx}\, dx \\ &= -\int_{-\infty}^\infty i e^{-x^2/2}(-ik)e^{-ikx}\,dx && \text{(integration by parts)}\\ &= -k \,\hat{\!f}(k). \end{align*} $$ This gives us the ordinary differential equation $\hat{\!f}'(k) = -k \,\hat{f}(k)$ with the initial condition $$ \hat{\!f}(0) = \int_{-\infty}^{\infty} e^{-x^2/2}\,dx = \sqrt{2\pi}, $$ which shows that $$\color{blue}{\hat{f}(k) = \sqrt{2\pi}\,e^{-k^2/2}}.$$

In order for you to be able to check your own work, here's my solution of the exercise (N. S. suggested a slightly different approach, but neither is simpler or more efficient, they come down to the same):

We have $f'(x) = -x\,f(x)$ and $f''(x) = (x^2-1)f(x)$ so that $$g(x) = (x+1)^2 f(x) = f''(x) -2f'(x) + 2f(x),$$ hence $$ \begin{align*} \hat{g}(k) &= \widehat{f''}(k) - 2\widehat{f'}(k) + 2\widehat{f}(k) \\ &= (-k^2+2ik+2) \hat{f}(k) \end{align*} $$ which gives us $$\color{red}{\hat{g}(k) = \sqrt{2\pi}\,(-k^2+2ik+2)\,e^{-k^2/2}},$$ where you can write $-k^2+2ik+2 = 1+(1+ik)^2$ if you prefer.

It remains to justify $(!!)$: Compute $$ \begin{align*} \frac{d}{dk}\hat{\!f}(k) & = \lim_{h\to 0} \frac{1}{h} \left(\hat{\!f}(k+h)-\hat{\!f}(k)\right) \\ &= \lim_{h\to0} \int_{-\infty}^{\infty} e^{-x^2/2} \frac{e^{-ikx}}{h}\underbrace{\left(e^{-ihx}-1\right)}_{\large 2i e^{ihx/2}\sin{\frac{hx}{2}}}\,dx. \end{align*} $$ Using that $|\sin{\frac{hx}{2}}| \leq \dfrac{|hx|}{2}$ we see that the absolute value of the integrand is bounded above by $|x|e^{-x^2/2}$ which is clearly integrable, so we conclude by the dominated convergence theorem.