Proving that an integral defines a rational function without evaluating

Question

For $x\in\mathbb{R}$ define $$I(x):=\int_0^\infty t e^{-t} \sin(xt) dt.$$ It can be shown that $$I(x)=\frac{2 x}{\left(x^2+1\right)^2}.$$

Question: Can we use a suitable characterization of rational functions to verify that $I(x)$ defines a rational function directly on the integral expression, without evaluating?