Deriving the integration property of the Fourier Transform
First, recognise the integral of $x(t)$
$$\int_{-\infty}^{t} x(\tau) d\tau \tag{1}$$
as a convolution with $u(t)$
$$ \int_{-\infty}^{t} x(\tau) d\tau =x(t) \star u(t) \tag{2}$$
where $u(t)$ is the unit-step function.
Then from the convolution property of the CTFT, one gets the following :
$$ \mathcal{F}\{ \int_{-\infty}^{t} x(\tau) d\tau  \} = \mathcal{F}\{ x(t)\star u(t) \} = X(\omega)U(\omega) \tag{3}$$ where $U(\omega)$ is the Fourier transform of the unit-step function which is $$U(\omega) = \frac{1}{j\omega} + \pi \delta(\omega) \tag{4}$$
Then the Fourier transform of the integral of $x(t)$ is
$$ \mathcal{F}\{ \int_{-\infty}^{t} x(\tau) d\tau \} = \frac{ X(\omega) }{j\omega} + \pi X(0) \delta(\omega) \tag{5}$$