广义积分∫(0,+∞) 1/(x^2+2X+3)dx为

题目

广义积分∫(0,+∞) 1/(x^2+2X+3)dx为
√2/2 *( pai/2-arctan√2/2)

答案

∫(0-->+∞) 1/(x²+2x+3)dx
=∫(0-->+∞) 1/(x²+2x+1+2)dx
=∫(0-->+∞) 1/((x+1)²+2)dx
=(1/√2)*arctan[(x+1)/√2] ∫(0-->+∞)
=(1/√2)*[π/2-arctan(1/√2)]