$\displaystyle \int_{-2}^{2} \frac{3dt}{4+3t^2} \, dt$
is my u=4+3t^2?
that won't work because, then $\displaystyle du=6t dt$.
Since there is no t on the upper side of the fraction (sorry I can't remember the name that is called) it will be a dead end.
try using the identity,
$\displaystyle \frac{d\tan{^{-1}}x}{dx}=\frac{1}{1+x^2}$