$ x(t) = e^{-3|t-2|} $
Noticing that there is an absolute value, we can proceed to divide in tow cases.
When
$ t-2 < 0 \rightarrow x_1(t) = e^{3t-6} $
and when,
$ t-2 >0 \rightarrow x_2(t) = e^{-3t-6} $
So, we can then compute the Fourier series by adding the integrals of each diferent case.
$ \ \mathcal{X}(\omega) = \int_{-\infty}^{\infty} x_1(t)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty} x_2(t)e^{-j\omega t} \,dt $
$ \mathcal{X}(\omega) = \int_{-\infty}^{2} e^{3t-6}e^{-j\omega t}\,dt + \int_{2}^{\infty} e^{-3t-6}e^{-j\omega t} \,dt $
$ \mathcal{X}(\omega) = \frac{1}{e^{6}} \int_{-\infty}^{2} e^{3t-j\omega t}\,dt + \frac{1}{e^{6}} \int_{2}^{\infty} e^{-3t-j\omega t} \,dt $
$ \mathcal{X}(\omega) = \frac{1}{e^{6}} \int_{-\infty}^{2} e^{t(3-j\omega)}\,dt + \frac{1}{e^{6}} \int_{2}^{\infty} e^{-t(3+j\omega)} \,dt $
$ {\left. \frac{e^{-(j\omega + 2)t}}{-(j\omega +2)} \right]^{\infty}_0 } \frac{1}{e^{6}} + {\left. \frac{e^{-(j\omega + 4)t}}{-(j\omega +4)} \right]^2_0 } \frac{1}{e^{6}}\, $
