The particle is falling down, and $x$ is measured in down direction.
$$x''=g-kv^4 \\ \implies\quad v\frac{dv}{dx}=g-kv^4$$
Solving this I get $$\frac{1}{4\sqrt{gk}}\log\bigg(\frac{\sqrt k v^2-\sqrt g}{\sqrt k v^2+\sqrt g}\bigg)=x+c$$
It is falling from rest, i.e $x=0, v=0$, using this initial condition I get $ \log(-1)$ on the LHS. Is something wrong with the solution. I check that integral is correct.
Hint: Your answer is wrong From $v\frac{dv}{dx}=g-kv^4$ we have $$\int\dfrac{v\ dv}{g-kv^4}=dx$$ let substitution $v^2=\sqrt{\frac{g}{k}}w$ then $$x=\dfrac{1}{2\sqrt{kg}}\int\dfrac{dw}{1-w^2}+C=\dfrac{1}{4\sqrt{kg}}\ln\dfrac{1+w}{1-w}+C$$ now let $x=0$, $v=0$ then $C=0$.