求∫(0到+无穷)1/x^2+4x+8dx=派/8

Question

如题所述

Answer 1

被积函数1/(x^2+4x+8)=1/((x+2)^2+2^2)，原函数是1/2×arctan((x+2)/2)，代入上下限，积分结果是1/2×π/2-1/2×π/4=π/8