The function $f(x)=\log_e(x^3+\sqrt{x^6+1})$ is

even and increasing odd and increasing even and decreasing odd and decreasing

odd and increasing

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.