Answer
$\dfrac{\pi}{16}$
Work Step by Step
Consider $x =2\tan \theta \implies dx= 2 \sec^2 \theta d \theta $
Then,
$\int \dfrac{2 \sec^2 \theta d \theta }{8 \sec^2 \theta} =\int (\dfrac{1}{4} ) d \theta$
and
$ (\dfrac{1}{4}) \int_{0}^{2} d \theta=[\dfrac{1}{4} \theta +C]_{0}^{2}$
Thus, $\dfrac{1}{4}[\tan^{-1} (1)-\tan^{-1} (0)]=\dfrac{\pi}{16}$
