If $I =\int\limits_{-1}^{1}(\frac{x^2+sinx}{1+x^2})dx$, then its value is

0 2 π/2 2 - π/2

2 - π/2

$I =\int\limits_{-1}^{1}\frac{x^2}{1+x^2}dx+0$  (because $\frac{sinx}{1+x^2}$ is odd) $I =2\int\limits_{0}^{1}\frac{x^2+1-1}{x^2+1}dx$ $=2\int\limits_{0}^{1}dx-2\int\limits_{0}^{1}\frac{dx}{1+x^2}=2-2tan^{-1}x|_0^1=2-\frac{\pi}{2}$ Hence (D) is the correct answer.