## Calculus (3rd Edition)

$$\frac{1}{4}\ln \frac{5}{3}.$$
Since $u=4t+12$, then $du=4dt$; moreover, when $t:0\to 2$, then $u:12\to 20$ and hence, we have $$\int_{0}^{2} \frac{d t}{4 t+12}= \frac{1}{4}\int_{12}^{20} \frac{d u}{u}=\frac{1}{4}\ln u|_{12}^{20}\\=\frac{1}{4}(\ln 20-\ln 12)=\frac{1}{4}\ln \frac{20}{12}=\frac{1}{4}\ln \frac{5}{3}.$$