Answer
$$\frac{1}{6} \tan^{-1}\frac{8}{3}$$
Work Step by Step
Since $u=\frac{2}{3}t$, then $du= \frac{2}{3}dt$, and hence when $t:0\to 4 $, then $u:0\to 8/3$. Now, we have
$$\int_ 0^4\frac{ dt}{4t^2+9}=\frac{1}{6}\int_0^{8/3} \frac{ du}{u^2+1}=\frac{1}{6} \tan^{-1}u|_0^{8/3}\\
=\frac{1}{6} \tan^{-1}\frac{8}{3}-\frac{1}{6} \tan^{-1}0=\frac{1}{6} \tan^{-1}\frac{8}{3}.$$
