判断函数f(x)=ln[x+(根号X^2+1)]的奇偶性

因为f(x)=ln[x+(x^2+1)^(1/2)]所以f(-x)=ln[-x+(x^2+1)^(1/2)]所以f(x)+f(-x)=ln[x+(x^2+1)^(1/2)]+ln[-x+(x^2+1)^(1/2)]=ln{[x+(x^2+1)^(1/2)]*[-x+(x^2+1)^(1/2)]}=ln[(x^2+1)-x^2]=ln1=0所以-f(x)=f(-x)所以函数f...