Answer
$\int_{-1}^{1}\frac{sin~x}{1+x^2}~dx = 0$
Work Step by Step
The function $~~f(x) = sin~x~~$ is an odd function while $~~g(x) = 1+x^2~~$ is an even function. Then $~~\frac{sin~x}{1+x^2}~~$ is an odd function.
Therefore:
$\int_{-1}^{1}\frac{sin~x}{1+x^2}~dx = 0$