The function f(x)=log(x+√(x^2+1)), is:

$f(-x)=\log(-x+\sqrt{x^2+1})$

$=\log\left(-x+\sqrt{x^2+1}×\frac{(x+\sqrt{x^2+1})}{(x+\sqrt{x^2+1})}\right)$

$f(-x)=\log(\frac{1}{x+\sqrt{x^2+1}})=-\log(x+\sqrt{x^2+1})$

so $f(-x)=-f(x)$ an odd function