∫[-2,2] (1+sinx)/(1+x^2)dx=∫[-2,2] 1/(1+x^2)dx+∫[-2,2] sinx/(1+x^2)dx而sinx/(1+x^2)是奇函数,所以在[-2,2]对称区间内积分为0所以只用求∫[-2,2] 1/(1+x^2)dx= arctanx |[-2,2]= 2arctan2...