Signal
$ X(\omega) = 6\delta (\omega - 5) + 3\pi \delta(\omega - 1) \! $
Inverse Fourier Transform
$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega )e^{j\omega t} d\omega \! $
$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} 6\delta (\omega -5)e^{j\omega t} d\omega + \frac{1}{2\pi} \int_{-\infty}^{\infty} 3\pi \delta (\omega -1)e^{j\omega t} d\omega \! $
We know that:
$ \int_{-\infty}^{\infty} \delta (\omega -t_0) e^{jwt} d\omega = e^{jt_0 t} \! $
Therefore:
$ x(t) = \frac{6}{2\pi }e^{5jt} + \frac{3\pi }{2\pi }e^{jt} \! $
$ x(t) = \frac{3}{\pi }e^{5jt} + \frac{3}{2}e^{jt} \! $
