凑微+分部积分+变量替换记I=∫ (1~+∞)arctanx/(x^2) dx =-∫ (1~+∞)arctanxd(1/x )=-(1/x)arctanx|(1,+∞)+∫ (1~+∞)1/[x(1+x^2)]dx=π/4+∫ (1~+∞)1/[x(1+x^2)]dx令1/x=t.则∫ (1,+∞)1/[x(1+x^2)]dx=∫(0~1)t/...