${\displaystyle \underset{1}{\overset{3}{}}}x\sqrt[3]{x^2-1}dx=$

$($A$.)\frac{16\sqrt[\phantom{2}]{2}}{3}$

$($B$.)6$

$($C$.)\frac{3}{7}(2^{\frac{7}{3}}-1)$

$($D$.)\frac{3}{4}{2}^{\frac{4}{3}}$

$($E$.)1$