= Solution
The ordinary <integral> fails because the simple real poles at $-2$ and $3$ produce nonintegrable logarithmic singularities. Partial fractions give
$$
\frac1{x^2-x-6}=\frac15\left(\frac1{x-3}-\frac1{x+2}\right).
$$
Closing above for $\omega>0$ and below for $\omega<0$, with symmetric semicircular indentations, yields
$$
I(\omega)=\frac{i\pi}{5}\operatorname{sgn}(\omega)
\left(e^{3i\omega}-e^{-2i\omega}\right).
$$
Taking real and imaginary parts,
$$
\operatorname{PV}\int\frac{\cos\omega x}{x^2-x-6}\,dx
=-\frac{\pi}{5}\operatorname{sgn}(\omega)(\sin3\omega+\sin2\omega),
$$
$$
\operatorname{PV}\int\frac{\sin\omega x}{x^2-x-6}\,dx
=\frac{\pi}{5}\operatorname{sgn}(\omega)(\cos3\omega-\cos2\omega).
$$
All <integrals> here run over $\mathbb R$.
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