With some intermediate derivation results I found the two integrals are exactly the same. But why?
\begin{align}
I=\int_0^\infty\frac{\sin x}{x + x^2}\mathrm dx=\int_0^\infty\frac{\pi-2\tan^{-1}x}{2 e^x}\mathrm dx
\end{align}
Both are equal to
\begin{align}
I&=\operatorname{Si}(1)\cos(1)-\operatorname{Ci}(1)\sin(1)+\frac{\pi}{2}\left(1-\cos(1)\right)\\
&\approx0.9493467025590832615920\ldots
\end{align}
where $\operatorname{Ci}$ and $\operatorname{Si}$ are cosine and sine integrals, respectively.
Note: The nominator in integrand is the double of Laplace transform of $\operatorname{sinc}$ function
$$\mathcal{L}\left[\frac{\sin t}{t}\right](x)=\cot^{-1}{\frac1x}=\frac{\pi}{2}-\tan^{-1}x$$
Best Answer
Their equivalency is established below\begin{align} &\int_0^\infty\frac{\pi-2\tan^{-1}x}{2 e^x}dx\\ =& \int_0^\infty \left(\frac\pi2-\tan^{-1}x\right)d(1-e^{-x}) \overset{ibp} = \int_0^\infty \frac{1-e^{-x}}{1+x^2}dx\\ = &\int_0^\infty (1-e^{-x})\left( \int_0^\infty \sin y\ e^{-xy}\ dy \right)dx\\ = & \int_0^\infty \sin y \left(\int_0^\infty e^{-xy}(1-e^{-x})dx\right)dy\\ =& \int_0^\infty\frac{\sin y}{y + y^2}dy \end{align}