Answer
$\dfrac{\pi}{16}$
Work Step by Step
Let us consider $x =2\tan \theta $
and $dx= 2 \sec^2 \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{2 \sec^2 \theta d \theta }{8 \sec^2 \theta} =\dfrac{1}{4} \int d \theta$
Now, integrate with limits:
$\int_{0}^{2} \dfrac{1}{4} d \theta=[\dfrac{1}{4} \theta +C]_{0}^{2}=\dfrac{1}{4}[\tan^{-1} (1)-\tan^{-1} (0)]=\dfrac{\pi}{16}$
