letx= tanudx=(secu)^2 dux=0, u=0x=+∞ , u=π/2∫(0->+∞) dx/√(1+x^2)=∫(0->π/2) (secu)^2 du/ secu=∫(0->π/2) secu du = [ln|secu + tanu| ]|(0->π/2)=lim(u->π/2) ln|secu + tanu|发散！
积分就没求对
