Fourier transform of $\frac{1}{\sqrt{1 + x^2}}$

As already stated it depends upon how you really define the Fourier transform. If you define it as an improper integral the question becomes non-trivial (otherwise it is just Plancherel). For $n\geq 1$ let $$ f_n(x) = f(x)\; {\rm \bf 1}_{[-n,n]}(x) = \frac{1}{\sqrt{1+x^2}} {\rm \bf 1}_{[-n,n]}(x) $$ where ${\rm \bf 1}_{[-n,n]}$ is the indicator function for the interval $[-n,n]$. Then $f_n$ is in both $L^1\cap L^2$ and the Fourier transform $$ \hat{f}_n(t) = \int_{-n}^n \frac{e^{itx}}{\sqrt{1+x^2}} dx $$ converges for every $t\neq 0$. This follows from partial integration: $$ \hat{f}_n(t) = \left[ \frac{e^{itx}}{it\sqrt{1+x^2}} \right]_{-n}^n + \int_{-n}^n \frac{\sin(tx)}{t} \frac{x}{\sqrt{1+x^2}^3} dx . $$ As $n\rightarrow \infty$, the first term goes to zero and the latter converges absolutely. Thus, you may define $\hat{f}(t)=\lim_n \hat{f}_n(t)$ for every $t\neq 0$ and the question is then if $\hat{f}\in L^2$?

The answer is yes: First note, that $f_n \rightarrow f$ in $L^2$ so $(f_n)_{n\geq 1}$ is Cauchy in $L^2$.

By Plancherel $\|\hat{f}_n-\hat{f}_m\|_2 = \|f_n-f_m\|_2$, so $(\hat{f}_n)_{n\geq 1}$ is also Cauchy in $L^2$, whence converges in $L^2$ to some function $g$.

Any convergent sequence in $L^2$ has, however, a subsequence that converges a.e. and we already know that for $t\neq 0$ (i.e. a.e.) the limit is $\hat{f}(t)$, so $\hat{f}=g$ (mod 0) is in $L^2$

Answering confusion in the comments about the Plancherel theorem:

A rough version of the theorem is

$\hat f\in L^2(\mathbb R)$ if and only if $f\in L^2(\mathbb R)$.

That's true, but it really needs to be stated more carefully. Because for example it's not clear what $\hat f$ even means given just $f\in L^2$. A more careful version starts here:

If $f\in L^1\cap L^2$ then $\hat f\in L^2$ and $\lVert\hat f\rVert_2=\lVert f\rVert_2$.

There there's no problem with the definition of $\hat f$, since $f\in L^1$. $\newcommand{\ft}{\mathcal F}$ At this point it's convenient to introduce the notation

$$\ft f=\hat f.$$

Now $\ft$ is defined on a dense subspace of $L^2$ and $\ft$ is an isometry (in the $L^2$ norm) on that dense subspace. Hence $\ft$ extends to an isometry $\ft:L^2\to L^2$. When people talk about $\hat f$ for $f\in L^2$ they're referring to that extension. (And now we're done with the problem in the question: Since $f\in L^2$ it follows that $\hat f\in L^2$.)

It follows easily that $\ft$ is its own inverse, almost:

If $f\in L^2$ then $\ft^2f(t)=f(-t)$.

Proof: The $L^1$ inversion theorem shows that this holds for, say, $f$ in the Schwartz space. Since the Schwartz space is dense and $\ft$ is continuous on $L^2$ (as is the operator $f(t)\mapsto f(-t)$) it holds for all $f\in L^2$.

In particular $\ft$ maps $L^2$ onto $L^2$: that is, $\ft$ is an isometric isomorphism, which is to say $\ft$ is unitary. (And in fact $\ft^4=I$, which leads easily to a decomposition of $L^2$ into four eigenspaces...)