Let $L\left(d\right)$ be the arc length of the curve $g\left(x\right)=3x^2$ from $x=0$ to $x=d$, where $d\ge 0$. Find an explicit expression for $L\left(d\right)$.

$L\left(d\right)=2d\sqrt{1+6d^2}+\ln (3d+\sqrt{1+6d^2})$

$L\left(d\right)=\frac{d}{2}\sqrt{1+36d^2}+\frac{1}{12}\ln (6d+\sqrt{1+36d^2})$

$L\left(d\right)=3d\sqrt{1+6d^2}+\ln (\frac{d}{3}+\sqrt{1+6d^2})$

$L\left(d\right)=\frac{d}{6}\sqrt{1+36d^2}+\frac{1}{24}\ln (\frac{d}{6}+\sqrt{1+36d^2})$