Answer
$\dfrac{\pi}{4}$
Work Step by Step
Let us consider $x =2\tan \theta $
and $dx= 2 \sec^2 \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{2 \sec^2 \theta d \theta }{\sqrt {4+4 \tan^2 \theta)}} $
or, $=\dfrac{1}{2} \int d \theta$
Now, integrate with limits:
$\int_{-2}^{2} \dfrac{1}{2} d \theta=\dfrac{1}{2}[\tan^{-1} (1)-\tan^{-1} (-1)]=\dfrac{\pi}{4}$
