I have a fourier transform which is
$$X(jω)=\frac{\cos(2ω)}{ω^2+ω+1}$$
and I am trying to calculate the value of the integral:
$$∫x(t)dt \ \ \ \ \ \ x \in (-\infty, \infty)$$.
I was thinking I could use Parseval's theorem and tried doing the integral of
$$\frac{1}{2\pi} \int \biggr |\frac{\cos(2ω)}{(ω^2+ω+1)}\biggr|^{2} d\omega$$
which comes out to roughly $1.789$ but after that I'm stuck. Would greatly appreciate any advice on how to proceed or if there is an easier way of doing this?
Your integral, if it has a finite value and $X$ is continuous at $ω=0$, as it obviously is, has the value $\sqrt{2\pi}X(0)$.
This follows from the Fourier pair being $$ X(jω)=\frac1{\sqrt{2\pi}}\int_{\mathbb R}x(t)e^{-jωt}\,dt\text{ and }x(t)=\frac1{\sqrt{2\pi}}\int_{\mathbb R}X(jω)e^{jωt}\,dω $$ as long as the integrals can be evaluated (see extensions of Fourier transform to tempered distributions).
The first formula evaluated at $ω=0$ results in $$ X(j0)=\frac1{\sqrt{2\pi}}\int_{\mathbb R}x(t)\,dt. $$ There is no need to evaluate any norm integrals.
You want the value of $$ \begin{align} &\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\cos^{2}(2\omega)}{(\omega^{2}+\omega+1)^{2}}d\omega \\ & = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\cos^{2}(2\omega)}{(w-e^{2\pi i/3})^{2}(w-e^{-2\pi i/3})^{2}}d\omega \\ & = \frac{1}{4\pi}\int_{-\infty}^{\infty}\frac{\cos(4\omega)+1}{(w-e^{2\pi i/3})^{2}(w-e^{-2\pi i/3})^{2}}d\omega \\ & = \Re \frac{1}{4\pi}\int_{-\infty}^{\infty}\frac{e^{4i\omega}+1}{(w-e^{2\pi i/3})^{2}(w-e^{-2\pi i/3})^{2}}d\omega. \end{align} $$ Now you can close the integral in the upper half plane, which gives a value of $$ \left.\Re\left(\frac{2\pi i}{4\pi}\frac{d}{d\omega}\frac{e^{4i\omega}+1}{(w-e^{-2\pi i/3})^{2}}\right)\right|_{\omega=e^{2\pi i/3}}. $$
