∫√{[(x+1)/(x-1)]+√[(x-1)/(x+1)]}dx
(x+1)/(x-1)＞0
x＞1或x＜-1
-1＜1/x＜1
设1/x=cosβ ,0＜β＜π
∫√{[(x+1)/(x-1)]+√[(x-1)/(x+1)]}dx
=∫√{[(1+1/x)/(1-1/x)]+√[(1-1/x)/(1+1/x)]}dx
=∫√{[(1+cosβ)/(1-cosβ)]+√[(1-cosβ)/(1+cosβ)]}d(1/cosβ)
=∫[cot(β/2)+tan(β/2)]d(1/cosβ)
=∫[cot(β/2)+tan(β/2)]d(1/cosβ)
=∫[cos(β/2)/sin(β/2)+sin(β/2)/cos(β/2)]d(1/cosβ)
=∫{1/[cos(β/2)sin(β/2)]}d(1/cosβ)
=2∫(1/sinβ)d(1/cosβ)
=-2∫[(1/sinβ)(cosβ)^(-2)]d(cosβ)
=2∫[(1/sinβ)(cosβ)^(-2)]sinβdβ
=2∫[(cosβ)^(-2)]dβ
=2tanβ+c
=2tanarccos(1/x)+c
=2sinarccos(1/x)/cosarccos(1/x)+c
=2xsinarccos(1/x)+c
=2x√{1-[cosarccos(1/x)]^2}+c
=2x√(1-1/x^2)+c
=2(x/|x|)√(x^2-1)+c