Note that the variable of integration is $t$, so, we can use the following steps.
$$\int_{0}^{\infty}e^{-(\alpha-\omega)t}\|x\|dt \Rightarrow\|x\|\int_{0}^{\infty}e^{-(\alpha-\omega)t}dt$$.
Here we do a little substitution and let $v = (\alpha - \omega)t$ and so we have $\frac{dv}{dt}=(\alpha - \omega) \Rightarrow \frac{dv}{(\alpha-\omega)} = dt$. Upon substituting, we see that
$$\|x\|\int_{0}^{\infty}e^{-(\alpha-\omega)t}dt \Rightarrow \|x\|\lim_{R \rightarrow \infty}{\frac{1}{(\alpha - \omega)}\int_{0}^{R}e^{-v}dv} \Rightarrow \|x\| \frac{1}{(\alpha - \omega)} \lim_{R \rightarrow \infty} {\int_{0}^{R}e^{-v}dv}$$
Which evaluates to $$\|x\| \frac{1}{(\alpha - \omega)}$$ Since $$\lim_{R \rightarrow \infty}{\int_{0}^{R}e^{-v}dv} = 1$$