I want to find the Fourier transform of $$f(x)=\frac{4}{3+2x+x^2}$$
We worked with the following definition:
$$\hat f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-itx}dt$$
Now, $f(x)=\frac{4}{(x+1)^2+2}=\frac{2}{(\frac{x+1}{\sqrt{2}})^2+1}$.
So $\hat f(x)=2\hat g(\frac{x+1}{\sqrt 2})$, $a=1$.
Now I struggle a bit to determine $\hat g(\frac{x+1}{\sqrt 2})$:
$\hat g(\frac{x+1}{\sqrt 2})=\sqrt 2 \frac{1}{\sqrt 2} \hat g(\frac{x+1}{\sqrt 2})=\sqrt 2 \hat g(x+1)=\sqrt 2 \hat g(x)e^{ix}=\sqrt 2 e^{ix}\frac{\sqrt{\frac{\pi}{2}}e^{-1|x|}}{1}$
Which means $\hat f(x)=2(...)$
Can you tell me if this is correct? And if you know better ways to do this sort of examples please let me know.
EDIT:
Your equation $\frac 1 {\pi} \int e^{-isx} \frac 1 {1+s^{2}}ds=e^{-|x|}$ is equivalent to
$\frac 1 {\sqrt{2\pi}} \int e^{-isx} \frac 1 {1+s^{2}}ds=\sqrt{\frac{\pi}{2}}e^{-|x|}$
The last equation means that the Fourier transform of $g(x):=\frac{1}{1+x^2}$ is $\hat g(x)=\sqrt{\frac{\pi}{2}}e^{-|x|}$.
We have $f(x)=\frac{4}{(x+1)^2+2}=\frac{2}{(\frac{x+1}{\sqrt{2}})^2+1}$.
So $\hat f(x)=2 \hat g(\frac{x+1}{\sqrt 2})$.
I hope everything is correct so far. But can you help me find my mistakes here:
$\hat g(\frac{x+1}{\sqrt 2})=\sqrt 2 \frac{1}{\sqrt 2} \hat g(\frac{x+1}{\sqrt 2})=\sqrt 2 \hat g(x+1)=\sqrt 2 \hat g(x)e^{ix}=\sqrt 2 e^{ix}\sqrt{\frac{\pi}{2}}e^{-|x|}$
In the 2nd equation I used the property $g(t)= |r|f(rt) \Rightarrow \hat g(x)=\hat f(x/r)$. And in the 4th I used $f_s(t):=f(t+s) \Rightarrow \hat f_s(x)=\exp(its)\hat f(x)$
Making the substitution $s=\frac {1+t} {\sqrt 2}$ will give the answer as $2\sqrt {\pi} e^{-x-\sqrt 2 |x|}$ if you use the formula $\frac 1 {\pi} \int e^{-isx} \frac 1 {1+s^{2}}ds=e^{-|x|}$.