Answer
$$\frac{1}{4}\ln\frac{5}{3}.$$
Work Step by Step
We have
$$\int_0^2 \frac{dt}{4t+12}=\frac{1}{4}\int_0^2 \frac{4dt}{4t+12}=\frac{1}{4}\ln (4t+12)|_0^2\\
=\frac{1}{4}(\ln 20 -\ln 12)=\frac{1}{4}\ln \frac{20}{12}=\frac{1}{4}\ln\frac{5}{3}.$$
