<s>[INST] Fourier transform of $f(x)=\frac{1}{x^2+6x+13}$ How to find the Fourier transform of the following function:
$$f(x)=\frac{1}{x^2+6x+13}$$
 [/INST] This begs for completing the square:
$$
x^2 + 6x + 13 = \Big( x^2 + 6x + 9 \Big) + 4 = (x+3)^2 + 4.
$$
Then
$$
\frac{1}{(x+3)^2 + 4} = \frac 1 4 \cdot \frac{1}{\left(\dfrac{(x+3)^2}{4}\right)  + 1} =\frac 1 4 \cdot \frac{1}{\left(\frac{x+3}{2}\right)^2+1} = \frac 1 4 \cdot \frac{1}{w^2 + 1}
$$
The Laplace transform involves an integral with respect to $x$.  So
$$
w = \frac{x+3}{2}
$$
$$
dw = \frac{dx}{2}
$$
$$
dx = 2\,dw
$$
If you can find $\displaystyle \int_0^\infty \frac{1}{1+w^2} e^{-tw}\,dw$, you should be able to get what you're looking for.

If you'd had a number smaller than $9$ where $13$ appeared, then things would be different.  For example:
$$
x^2 + 6x + 5 = \Big(x^2+6x+0\Big) - 4 = (x+3)^2 - 2^2 = (x+3+2)(x+3-2)
$$
$$
=(x+5)(x+1).
$$
Then
$$
\frac{1}{x^2+6x+5} = \frac{1}{(x+5)(x+1)} = \frac{A}{x+5}+\frac{B}{x+1},
$$
and you'd have to find $A$ and $B$.
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