Answer
$\dfrac{3\pi}{4}$
Work Step by Step
Consider $f(x)= \int_{-\infty}^{2} \dfrac{2}{x^2+4} dx$
Since, we have $\lim\limits_{a \to -\infty} f(x)= \lim\limits_{a \to -\infty}\int_{a}^{2} \dfrac{2}{4[(x/2)^2+1]} dx=\lim\limits_{a \to -\infty}\int_{a}^{-2}[\tan^{-1} (1) -\tan^{-1} (a/2)] $
or, $=\lim\limits_{a \to -\infty} \dfrac{\pi}{4} - \tan^{-1} (a/2)$
or, $=\dfrac{\pi}{4} - (-\dfrac{\pi}{2})$
or, $=\dfrac{3\pi}{4}$
