The function f(x)=log_e(x^3+√(x^6+1)) is

We have, $f(x) + f(−x) = \log 1= 0⇒f (−x) = −f (x)$

So, f(x) is an odd function

Now, $f(x)=\log_e(x^3+\sqrt{x^6+1})$

$⇒f'(x)=\frac{1}{x^3+\sqrt{x^6+1}}\left(3x^2+\frac{6x^5}{2\sqrt{x^6+1}}\right)=\frac{3x^2}{\sqrt{x^6+1}}>0$

⇒ f(x) is increasing.