If $I=\int\limits_{-1}^1\left(\frac{x^2+\sin x}{1+x^2}\right) d x$ then

$2-\frac{\pi}{2}$

$I=\int\limits_{-1}^1 \frac{x^2}{1+x^2} d x+0$         (because $\frac{\sin x}{1+x^2}$ is odd)

$I=2 \int\limits_0^1 \frac{x^2+1-1}{x^2+1} dx=2 \int\limits_0^1 d x-2 \int\limits_0^1 \frac{d x}{1+x^2}=2-\left.2 \tan ^{-1} x\right|_0 ^1=2-\frac{\pi}{2}$

Hence (4) is the correct answer.