设x=secu,2π/3<=u<=3π/4，dx=-sinudu/(cosu)^2,√(x^2-1)=-tanu,所以原式=∫<2π/3,3π/4>-du/cosu=-ln|secu+tanu||<2π/3,3π/4>=-[ln|-√2-1|-ln|-2-√3|]=ln|(√2-1）(2+√3)|=ln(2√2-2+√6-√3).
