f(x)=ln[x＋(√x^2＋1)]
f(-x)=ln[(-x)＋(√(-x)^2＋1)]
=ln[-x+(√x^2+1)]
=ln{[-x+(√x^2+1)][x+(√x^2+1)]/[x+(√x^2+1)]}
=ln[1/[x+(√x^2+1)]
=-ln[x+(√x^2+1)]=-f(x)
所以f(x)+f(-x)=0
f(x)是奇函数