your integral = $\displaystyle 2\int_0^{\infty} \frac{x \sin x}{x^2 + 1} \ dx - 2\int_0^{\infty} \frac{x \sin x}{(x^2 + 1)^2} \ dx,$ and by parts in the second integral gives us: your integral = $\displaystyle \int_0^{\infty} \frac{x \sin x}{x^2 + 1} \ dx=\int_0^{\infty} \frac{\cos x}{x^2 + 1} \ dx$why?

and i guess i don't have to prove this very well-known fact that $\displaystyle \int_0^{\infty} \frac{\cos x}{x^2 + 1} \ dx = \frac{\pi}{2e}.$