$23$:$12$

${\displaystyle \underset{0}{\overset{2}{}}}-6u[e^u-e^{\frac{u}{2}}]du$

We can make the substitution $z=e^u,$ which gives $dz=e^{u}du.$

The limits of integration for $z$ are $z=1$ and $z=e^2.$

The integral becomes:

$-6{\displaystyle \underset{1}{\overset{e}{}}}{?}^{{?}^{?}}2zlnzdz$

We can use integration by parts with $u=lnz$ and $dv=zdz.$

The integral becomes:

$\{e^{2}ln{e}^2$