$\int\limits_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x=$
Answer & explanation
Correct answer: option 3
$\int\limits_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x=\int\limits_{-1}^1 \frac{\sin x}{3-|x|} d x-\int\limits_{-1}^1 \frac{-x^2}{3-|x|} d x=0-\int\limits_0^1 \frac{2 x^2}{3-|x|} d x$
[∵ first integrand is an odd function and second is an even function.]
Hence (3) is the correct answer.