How find this $I=\int_{0}^{\infty}\frac{x\sin{(2x)}}{x^2+4}dx$

Find this integral
$$I=\int_{0}^{\infty}\dfrac{x\sin{(2x)}}{x^2+4}dx$$

let $x=2t$,
then
$$I=\int_{0}^{\infty}\dfrac{t\sin{(4t)}}{(t^2+1)}dt$$
then
$$I=1/2\int_{0}^{\infty}\sin{(4t)}d\ln{(t^2+1)}$$
then I can’t.

This problem have without residue methods?

Solutions Collecting From Web of "How find this $I=\int_{0}^{\infty}\frac{x\sin{(2x)}}{x^2+4}dx$"

This is a duplicate of this Functions defined by integrals (problem 10.23 from Apostol's Mathematical Analysis)

We have that
$$
F(y) = \int \frac{\sin(xy)}{x(x^2+1)}\mathrm{d}x
= \frac{\pi}{2}(1-e^{-y})
$$
A generalization of your integral is given as
$$
G(a,y) = \int \frac{\sin(xy)}{x(x^2+a^2)}\mathrm{d}x
= a^{-2} F(ay)
= \frac{\pi}{2a^2}(1-e^{-ay})
$$
Differentiating twice yield
$$
\frac{\mathrm{d}^2 G}{\mathrm{d}y^2}
= -\int_0^{\infty} \frac{x\sin(xy)}{x^2+a^2}
= – \frac{\pi}{2} e^{-ay}
$$
Hence
$$
\int_{0}^{\infty}\dfrac{x\sin{(2x)}}{x^2+4}\mathrm{d}x
= -G”(2,2)
= \frac{\pi}{2} e^{-4}
$$
Where $G”(a,y)$ means differentiation with respect to $y$.