题目

广义积分∫ [1/(x^2+4x+5)]dx = .

提问时间：2021-02-28

答案

∫<-∞,+∞> [1/(x²+4x+5)]dx
= ∫<-2,+∞> 1/[(x+2)²+1]d(x+2) + ∫<-∞,-2> 1/[(x+2)²+1]d(x+2)
=arctan(x+2)|<-2,+∞> +arctan(x+2)|<-∞,-2>
=π/2-0+0-(-π/2)
= π