不定积分——∫dx⼀(x*(x^2-1)^(1⼀2))

∫ dx/[x√(x²-1)]，令x = secy，dx = secy*tany dy
√(x²-1) = √(sec²y-1) = √(tan²y) = tany...(1) 或 -tany...(2)
siny = √(x²-1) / x，cosy = 1/x
若x>1，则0≤y<π/2
原式= ∫ secy*tany / [secy*(tany)] dy，第(1)种情况
= ∫ dy = y + C
= arcsec(x) + C

若x<-1，π/2原式= ∫ secy*tany / [secy*(-tany)] dy，第(2)种情况
= -y + C
= -arcsecx + C

对于x<-1，π/2设x = -z，z>1，dx = -dz
原式= ∫ -dz/[(-z)√(z²-1)]
= ∫ dz/[z√(z²-1)]，z = secy
= arcsec(z) + C
= arcsec(-x) + C
= π - arcsecx + C
= -arcsecx + C''，C''=π+C

综合∫ dx/[x√(x²-1)]有两个答案，分别是：
= arcsec(x) + C，当x>1
= -arcsecx + C，当x<-1

∫dx/[x√(x^2-1)]
=∫dx/[x^2√(1-1/x^2)]
=-∫d(1/x)/√[1-(1/x)^2]
t=1/x
=∫-dt/√(1-t^2)
=arccost +C
=arccos(1/x)+C

=arctan((x^2 - 1)^(1/2))+C