判断y=ln[x+(√x^2+1)]的奇偶性?

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判断y=ln[x+(√x^2+1)]的奇偶性?
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判断y=ln[x+(√x^2+1)]的奇偶性?
判断y=ln[x+(√x^2+1)]的奇偶性?

判断y=ln[x+(√x^2+1)]的奇偶性?
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