Even or Odd?!

$f\left( x \right) =\log { (x+\sqrt { { x }^{ 2 }+1 } } )\\ f\left( -x \right) =\log { (-x+\sqrt { { x }^{ 2 }+1 } } )\\ f\left( x \right) +f\left( -x \right) =\log { ({ x }^{ 2 } } +1-{ x }^{ 2 })=\log { (1)=0 } \quad \\ f\left( x \right) +f\left( -x \right) =0\quad \\ \boxed { f\left( -x \right) =-f\left( x \right) } \\$

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

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

Let us consider the term inside the $\log$ . $-x + \sqrt{x^2 +1}$

If we multiply this by $\frac{x + \sqrt{x^2 +1}}{x + \sqrt{x^2 +1}}$ , we get

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

Going back to the $\log$ , we now have:

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

Applying the law on logarithm

$\log M^n = n\cdot\log M$

we now have

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

Thus $f(-x) = -f(x)$ and $f$ is odd

if we take x=2(assume) then, f(2)=log(2+(4+1)^(1/2))=0.62696292,
f(-2)=log(-2+(4+1)^(1/2))= -0.62696292
so, f(x)=-f(x).. odd function