Finding Fourier transform of inverse polynomial I'm asked to find Fourier transform of $ f(t) = \frac{1}{t^2 + 2t + 2} $. I have searched for steps for solving this kind of problem (given a inverse polynomial) but couldn't find any. Should I factorize it to $ \frac{1}{(t+1-i)(t+1+i)} $ or? The given correct answer is $ F(w) = \pi e^{iw-|w|} $. 
If there is any similar example with solution, feel free to drop link and I can study the steps.
 A: Using the Fourier transform of $e^{- \alpha \lvert t \rvert}$, we find
\begin{align}
F(\omega) &= \frac{2 \alpha}{\alpha^{2} + \omega^{2}}
\end{align}
Hence,
\begin{align}
e^{- \alpha \lvert t \rvert} &= \frac{1}{2 \pi} \int_{\mathbb{R}} \frac{2 \alpha}{\alpha^{2} + \omega^{2}} e^{i \omega t} d \omega \\
\implies \pi e^{- \alpha \lvert t \rvert} &= \int_{\mathbb{R}} \frac{ \alpha}{\alpha^{2} + \omega^{2}} e^{i \omega t} d \omega \\
\end{align}
Now, set $\alpha = 1$ and map $t \mapsto -t$
\begin{align}
\implies \pi e^{- \lvert -t \rvert} &= \pi e^{- \lvert t \rvert} \\
&= \int_{\mathbb{R}} \frac{ 1}{1 + \omega^{2}} e^{- i \omega t} d \omega \\
\end{align}
Now map $\omega \mapsto \omega + 1$
\begin{align}
\implies \pi e^{- \lvert t \rvert} &= \int_{\mathbb{R}} \frac{ 1}{1 + (\omega + 1)^{2}} e^{- i (\omega + 1) t} d \omega \\
&= e^{-i t} \int_{\mathbb{R}} \frac{ 1}{1 + (\omega + 1)^{2}} e^{- i \omega t} d \omega \\
\implies \pi e^{- \lvert t \rvert + i t} &= \int_{\mathbb{R}} \frac{ 1}{1 + (\omega + 1)^{2}} e^{- i \omega t} d \omega
\end{align}
hence, by the Fourier inversion formula
\begin{align}
F(\omega) &= \int_{\mathbb{R}} \frac{ 1}{1 + (t + 1)^{2}} e^{- i \omega t} dt \\
&= \pi e^{i \omega - \lvert \omega \rvert} 
\end{align}
