令t=1/x,原式=∫dt/√ (4t^2+1)=1/2*∫d(2t)/√ ((2t)^2+1)=1/2ln(2t+√ (4t^2+1))+C=1/2ln(2/x+√ (4/x^2+1))+C
漏了负号？
