Answer
$$
\int \frac{ td t}{t^2+4}=\frac{1}{2}\ln |t^2+4|+c.
$$
Work Step by Step
Le $ u=t^2+4$ and hence $ du=2tdt $, then we have
$$
\int \frac{ td t}{t^2+4}=\frac{1}{2}\int \frac{ d u}{u}= \frac{1}{2}\ln |u|+c=\frac{1}{2}\ln |t^2+4|+c.
$$