Consider
$$f\left(z\right)=\frac{\left(z-1\right)e^{2iz}}{z^2-4z+5}$$
The poles are
$$z=2\pm i$$
$$\int _{\gamma }f(z)\,dz=2i\pi \operatorname{Res}\left(f,2+i\right)$$
$$\operatorname{Res}\left(f,2+i\right)=\lim _{z\to 2+i}\left(\frac{\left(z-1\right)e^{2iz}}{z-2+i}\right)$$
Which is $$\frac{\left(1-i\right)e^{4i}e^{-2}}{2i}$$
The integral over the half circle vanishes as $R\rightarrow +\infty$ so
$$\int _{-R}^{+R}\frac{\left(t-1\right)\left(\cos\left(2t\right)+i\sin\left(2t\right)\right)}{t^2-4t+5}\,dt=\frac{\pi \left(1+i\right)e^{4i}}{e^2}$$
Equating real and imaginary parts we finally have
$$\int _{-\infty}^{+\infty}\frac{\left(x-1\right)\cos(2x)}{x^2-4x+5}\,dx=\frac{\pi e^{4i}}{e^2}$$
That is to say
$$\frac{\pi \left(\cos(4)+i\sin(4)\right)}{e^2}$$
Here's the thing, Wolfram alpha says it's $$\frac{\pi \left(\cos(4)-\sin(4)\right)}{e^2}$$
What happened there?
You incorrectly equated real and imaginary parts; $$\frac{\pi(1+i)e^{4i}}{e^2}=\frac{\pi(1+i)(\cos{(4)}+i\sin{(4)})}{e^2}=\frac{\pi(\cos{(4)}-\sin{(4)})}{e^2}+\frac{\pi(\cos{(4)}+\sin{(4)})}{e^2}i$$ so the real part is $$\frac{\pi(\cos{(4)}-\sin{(4)})}{e^2}$$