Question

反常积分问题

反常积分问题第四题，答案是π/4，但是我自己算了两遍还是π/2。求解

Answer 1

应该是这样吧？不对的话再交流

Answer 2

Answer 3

(4) I = ∫<0, +∞> dx/[(1+x)(1+x^2)]
= (1/2)∫<0, +∞> [1/(1+x) + (1-x)/(1+x^2)]dx
= (1/2)∫<0, +∞> [1/(1+x) + 1/(1+x^2) - x/(1+x^2)]dx
= (1/2)[ln(1+x) + arctanx - (1/2)ln(1+x^2)]<0, +∞>
= (1/2) [ln{(1+x)/√(1+x^2)} + arctanx]<0, +∞>
= (1/2)[lim<x→+∞>ln{(1+x)/√(1+x^2)} + π/2]
= (1/2)[lim<x→+∞>ln{(1/x+1)/√(1/x^2+1)} + π/2] = π/4