If I=∫limits_-1^1≤ft((x^2+sin x)/(1+x^2)) d x then

$I=\int\limits_{-1}^1 \frac{x^2}{1+x^2} d x+0$         (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.