设函数 $f(x, y)=1+\frac{x y}{\sqrt{1+y^3}}$, 则积分 $I=\int_0^1 \mathrm{~d} x \int_{x^2}^1 f(x, y) \mathrm{d} y=$
$\text{A.}$ $\frac{1}{3}(\sqrt{2}+1)$.
$\text{B.}$ $\frac{1}{6}(\sqrt{2}-1)$.
$\text{C.}$ $\frac{1}{6}(\sqrt{2}+1)$.
$\text{D.}$ $\frac{1}{3}(\sqrt{2}-1)$.