Question

Find $f\left(x\right)=\log (x+\sqrt{1+x^2})$ is even or odd.

Answer 1

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

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

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

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

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

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

Thus $f\left(x\right)$ is an odd function.