求根号（x^2-1）的对x不定积分

∫√（x^2-1）dx
令x=sect
则
∫√（x^2-1）dx
=∫tantdsect=∫tan^2tsectdt=∫(sec^2t-1)sectdt=∫(sec^3t-sect)dt
=tant*sect-∫sec^3tdt
即
∫(sec^3t-sect)dt=tant*sect-∫sec^3tdt
2∫(sec^3t)dt=tant*sect+∫sectdt
∫sec^3tdt=1/2tant*sect+1/2ln|sect+tant|+c
所以
∫√（x^2-1）dx=tant*sect-∫sec^3tdt
=1/2tant*sect-1/2ln|sect+tant|+c
=1/2x√（x^2-1）-1/2ln|x+√（x^2-1）|+c