令√(x^2-1)=t,所以x=√(t^2+1),dx=t/√(t^2+1)dt所以∫√(x^2-1)/xdx=∫t/√(t^2+1)*[t/√(t^2+1)]dt=∫t^2/(t^2+1)dt=∫dt-∫1/(t^2+1)dt=t-arctant+C将t=√(x^2-1)代人可得∫√(x^2-1)/xdx=√(x^2-1)-arctan√(x^...