For all real numbers $x,$ the function $f(x)$ satisfies

$6+f(x)=2f(-x)+3x^2({\displaystyle \underset{-1}{\overset{1}{}}}f(t)dt)$

It follows that ${\displaystyle \underset{-1}{\overset{1}{}}}f(x)dx$ equals

$($A$)4$

$($B$)6$

$($C$)11$

$($D$)\frac{27}{2}$