Chọn phương án C.

Ta có \(-|x|=\begin{cases}
-x &\text{khi }x\geq0\\
x &\text{khi }x<0.
\end{cases}\)

Phương trình hoành độ giao điểm $$\begin{aligned}
x^2-2=-|x|\Leftrightarrow&|x|=2-x^2\\
\Leftrightarrow&\left[\begin{array}{l}x=2-x^2\\ x=x^2-2\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x^2+x-2=0\\ x^2-x-2=0\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.
\end{aligned}$$

Vậy ta có $$\begin{aligned}
S&=\displaystyle\int\limits_{-1}^{1}\left|x^2-2+|x|\right|\mathrm{\,d}x\\
&=\displaystyle\int\limits_{-1}^{0}\left|x^2-2-x\right|\mathrm{\,d}x+\displaystyle\int\limits_{0}^{1}\left|x^2-2+x\right|\mathrm{\,d}x\\
&=\dfrac{7}{6}+\dfrac{7}{6}=\dfrac{7}{3}.
\end{aligned}$$