Answer
$2 \pi$
Work Step by Step
Consider $I= \int_{-1}^{1} \int_{-1/\sqrt {1-x^2}}^{1/\sqrt {1-x^2}} (2y+1) dy dx$
or, $= \int_{-1}^{1}[y^2+y] dx$
or, $=[2 \sin^{-1} x] _{-1}^{1}$
Thus, we have $I=2 \arcsin (1)-2 \arcsin (-1) =\pi+\pi=2 \pi$
