Answer
$\frac{\sqrt 3}{2}-1+\frac{\pi}{6}$
Work Step by Step
We have
$$\int_{-1/2}^0\frac{(x+1)dx}{\sqrt{1-x^2}}=\int_{-1/2}^0\frac{x dx}{\sqrt{1-x^2}}+\int_{-1/2}^0\frac{dx}{\sqrt{1-x^2}}\\
=-\frac{1}{2}\int_{-1/2}^0\frac{-2x dx}{\sqrt{1-x^2}}+\int_{-1/2}^0\frac{dx}{\sqrt{1-x^2}}\\
=-\frac{1}{2}\frac{ \sqrt{1-x^2}}{1/2}|_{-1/2}^0+\sin^{-1}x|_{-1/2}^0\\
=-1+\frac{\sqrt 3}{2}+0+\frac{\pi}{6}\\=\frac{\sqrt 3}{2}-1+\frac{\pi}{6}
.$$
