Answer
$\dfrac{\pi}{6}$
Work Step by Step
Here, we have $\int_{2}^{4} \dfrac{dt}{t \sqrt {t^2-4}}=\lim\limits_{a \to 2^{+}}\int_{a}^{4} \dfrac{dt}{t \sqrt {t^2-4}}$
This implies that
$\lim\limits_{a \to 2^{+}}\int_{a}^{4} \dfrac{dt}{t \sqrt {t^2-4}}=\lim\limits_{a \to 2^{+}}[(\dfrac{1}{2})\sec^{-1} \dfrac{t}{2}]_{b}^{4}$
Thus, $\lim\limits_{a \to 2^{+}}(\dfrac{1}{2})\sec^{-1} \dfrac{4}{2}-\lim\limits_{a \to 2^{+}}(\dfrac{1}{2})\sec^{-1} \dfrac{a}{2}=\dfrac{\pi}{6}$
