= Solution
Splitting the <Fourier transform> integral at zero gives two absolutely convergent <integrals>:
$$
\begin{aligned}
\widehat{g_t}(\xi)
&=\int_0^\infty e^{-(t+i\xi)x}\,dx+\int_0^\infty e^{-(t-i\xi)x}\,dx\\
&=\frac1{t+i\xi}+\frac1{t-i\xi}
=\boxed{\frac{2t}{t^2+\xi^2}}.
\end{aligned}
$$
Both $g_t$ and $\widehat{g_t}$ are <integrable functions>. Let $h_t(x)=(t^2+x^2)^{-1}$, so $\widehat{g_t}=2t h_t$. The <duality property of the Fourier transform>, or <Fourier inversion> in <tempered distributions>, gives
$$
2t\widehat{h_t}(\xi)=\widehat{\widehat{g_t}}(\xi)=2\pi g_t(-\xi).
$$
These <Fourier transforms> have continuous representatives by the <Riemann-Lebesgue lemma>, so the identity holds at every $\xi$. Since $g_t$ is an <even function>,
$$
\boxed{\mathcal F_{x\to\xi}\!\left(\frac1{t^2+x^2}\right)=\frac\pi t e^{-t|\xi|}.}
$$
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