Answer
$\dfrac{\pi}{16}$
Work Step by Step
Our aim is to integrate the integral as follows:
$\int^2_1 \int^{\sqrt{2x-x^2}}_0 \dfrac{1}{(x^2+y^2)^2} \space dy \space dx =\int^{\pi/4}_0 \int^{2 \cos\theta}_{\sec\theta}\dfrac{1}{r^4} \space dr \space d\theta $
or, $=\int^{\pi/4}_0[\dfrac{-1}{2r^2}]^{2 \cos\theta}_{\sec\theta} \space d\theta $
=$\int^{\pi/4}_0(\dfrac{1}{2} \cos^2\theta-\dfrac{1}{8} \sec^2\theta) d\theta $
or, $=[\dfrac{1}{4}\theta+\dfrac{1}{8} \sin2\theta-\dfrac{1}{8} \tan\theta]^{\pi/4}_0$
or, $=\dfrac{\pi}{16}$
