Fourier Transform
$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $
$ x(t)=(t-1)e^{-6t+6}u(t-1) \,\ $
$ X(\omega)=\int_{-\infty}^{\infty}x(t)=(t-6)e^{-6t+6}u(t-6) e^{-j\omega t}dt \; $
$ x(t) \,\ $looks like $ te^{-6t}u(t) \,\ $ so we evaluate that
the F.T of $ te^{-6t}u(t) \,\ $ is
$ \int_{-\infty}^{\infty}te^{-6t}u(t) e^{-j\omega t}dt \; $
$ =\int_{0}^{\infty}te^{-6t-j\omega t}dt \; $
$ =\int_{0}^{\infty}te^{-t(6+j\omega t)}dt \; $
Do integration by parts
$ ={\left. \frac{-te^{-t(6+j\omega )}}{6+j\omega }\right]_{0}^{\infty}} - \int_{0}^{\infty}\frac{-te^{-t(6+j\omega )}}{6+j\omega }dt \; $
$ ={\left. \frac{-e^{-t(6+j\omega )}}{(6+j\omega)^2 }\right]_{0}^{\infty}} $
$ = \frac{1}{(6+j\omega)^2} $
And now we use the time shift property and get
$ X(\omega)=\frac{e^{-j\omega}}{(6+j\omega)^2} $
