Answer
$$2 \pi$$
Work Step by Step
We must integrate the integral as follows:
$$ \iint_{R} f(x,y) dA = \int_{-1}^{1} \int_{-1/\sqrt {1-x^2}}^{1/\sqrt {1-x^2}} (2y+1) dy dx \\= \int_{-1}^{1}[y^2+y] \space dx \\=2 \times [\sin^{-1} x] _{-1}^{1} \\=2 [\sin^{-1} (1)-\sin^{-1} (-1)] \\=\pi+\pi \\=2 \pi$$
