ln[x-√(1+x^2)] '
= 1/[x-√(1+x^2)] * [x-√(1+x^2)] '
= 1/[x-√(1+x^2)] * [1- x/√(1+x^2)]
= 1/[x-√(1+x^2)] * [√(1+x^2) -x] / √(1+x^2)
= -1/√(1+x^2)
同理,
ln[x-√(1-x^2)] '
= 1/[x-√(1-x^2)] * [x-√(1-x^2)] '
= 1/[x-√(1-x^2)] * [1+ x/√(1-x^2)]
= [√(1-x^2) +x] / [x*√(1-x^2) -1+x^2]