Answer
$\dfrac{3\pi}{4}$
Work Step by Step
Let $f(x)= \int_{-\infty}^{2} \dfrac{2}{x^2+4} dx$
Now, $\lim\limits_{k \to -\infty} f(x)= \lim\limits_{k \to -\infty}\int_{k}^{2} \dfrac{2}{4[(x/2)^2+1]} dx=\lim\limits_{k \to -\infty}\int_{k}^{-2}[\tan^{-1} (1) -\tan^{-1} (\dfrac{k}{2})] $
Then, $ \dfrac{\pi}{4} -\lim\limits_{k \to -\infty}[ \tan^{-1} (\dfrac{k}{2})]=\dfrac{\pi}{4} - (-\dfrac{\pi}{2})=\dfrac{3\pi}{4}$
