Compute the Fourier Transform of x(t):

$\,x(t)=2e^{-3t}u(t)+4[u(t+3)-u(t-3)]$

Using the Formula for Fourier Transforms:

$\mathcal{F}(x(t))= \mathcal{X}(\omega)= \int_{-\infty}^{\infty}x(t)e^{-j\omega t} \,dt$

So the calculation follows as:

$\mathcal{X}(\omega)= \int_{-\infty}^{\infty}(2e^{-3t}u(t)+4[u(t+3)-u(t-3)])e^{-j\omega t} \,dt$

$=2\int_{0}^{\infty}e^{-t(3+j\omega)}\,dt + 4\int_{-3}^{3}e^{-j\omega t}\, dt$

$=2\frac{-e^{-t}}{3+j\omega}\bigg|^{\infty}_{0}+4\frac{-e^{-j\omega t}}{j\omega}\bigg|^{3}_{-3}$

$=2\frac{1}{3+j\omega}+4*\frac{2}{\omega}*\frac{-e^{-3j\omega} + e^{3j\omega}}{2j}$ and using eulers identity for sin:

$=\frac{2}{3+j\omega}+\frac{8sin(3\omega)}{\omega}$