Answer
$$\frac{\pi}{6}$$
Work Step by Step
\begin{align*}
\int_{2}^{4} \frac{d t}{t \sqrt{t^{2}-4}}&=\lim _{b \rightarrow 2^{+}}\int_{b}^{4} \frac{d t}{t \sqrt{t^{2}-4}}\\
&=\lim _{b \rightarrow 2^{+}}\left[\frac{1}{2} \sec ^{-1} \frac{t}{2}\right]\bigg|_{b}^{4}\\
&=\lim _{b \rightarrow 2^{+}}\left[\left(\frac{1}{2} \sec ^{-1} \frac{4}{2}\right)-\frac{1}{2} \sec ^{-1}\left(\frac{b}{2}\right)\right]\\
&=\frac{1}{2}\left(\frac{\pi}{3}\right)-\frac{1}{2} \cdot 0=\frac{\pi}{6}
\end{align*}
