Answer
$2+\dfrac{\pi}{2}$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{0}^{2} \dfrac{x}{\sqrt {4-x^2}}+\int_{0}^{2} \dfrac{1}{\sqrt {4-x^2}}=$ ....(1)
Plug $u(\theta)=4-x^2 \implies du=-2xdx$
Then, we have
$\sin^{-1} (x/2) dx|_0^2+(\dfrac{-1}{2} \int_{0}^{4}\dfrac{du}{\sqrt u}=\sin^{-1}(1)-\sin^{-1}(0)- (\dfrac{1}{2}) (2) u^{1/2}$
Thus, equation (1) becomes
$\dfrac{\pi}{2}-0-(-2)=2+\dfrac{\pi}{2}$
