$\int\limits_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x=$

$\int\limits_0^1 \frac{-2 x^2}{3-|x|} d x$

$\int\limits_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x=\int\limits_{-1}^1 \frac{\sin x}{3-|x|} d x-\int\limits_{-1}^1 \frac{-x^2}{3-|x|} d x=0-\int\limits_0^1 \frac{2 x^2}{3-|x|} d x$

[∵ first integrand is an odd function and second is an even function.] 

Hence (3) is the correct answer.