Answer
$\dfrac{-2(x+4) \sqrt {2-x}}{3}+C$
Work Step by Step
Using $u$-substitution, we get:
$u^2=2-x \implies -dx= 2u \space du$
We integrate the integral as follows:
$I=\int \dfrac{(2-u^2)(-2u \space du)}{(u^2)^{1/2}}=\int 2u^2-4 du$
Now, $I=\dfrac{2u^3}{3}-4u+C \\=\dfrac{2(2-x) \sqrt {2-x}}{3}-4 \sqrt {2-x}+C \\=\dfrac{-2(x+4) \sqrt {2-x}}{3}+C$
