integration - Evaluating $int_{-infty}^infty frac{sin x}{x(x^2 +1)}dx$

When constructing appropriate contours, we would like it so that the singularities are not on the contour but rather inside or outside the contour.

I see that the integrand has a removeable singularity at $z=0$. Does this matter? I feel like even if we define $f(z) = 1$, then $f$ is analytic at $z=0$, so it is fine to construct an integral passing through $z=0$, (and it will never be fine to construct one passing through $\pm i$.)

Answer

Yes, you are right. If we define$$f(z)=\begin{cases}\dfrac{\sin z}z&\text{ if }z\neq0\\1&\text{ otherwise,}\end{cases}$$then your integral is$$\int_{-\infty}^{+\infty}\frac{f(x)}{x^2+1}\,\mathrm dx.$$Since $f$ is an analytic function, you can use the standard methods of Complex Analysis.