Answer
$$\int \sqrt{3-2x-x^{2}}dx=2arcsin(\frac{x+1}{2})+\frac{x+1}{2}\sqrt{3-2x-x^{2}}+C$$
Work Step by Step
$$\int \sqrt{3-2x-x^{2}}dx=\int \sqrt{4-(x+1)^{2}}dx$$
Let:
$$\ x+1=2sin\,t,\,dx=2cos\,tdt,$$
$$t=arcsin(\frac{x+1}{2}),cost=\frac{\sqrt{3-2x-x^{2}}}{2}$$
$$\int \sqrt{3-2x-x^{2}}dx=2\int \sqrt{4-4sin^{2}t}cos\,tdt$$
$$=4\int cos^{2}t\,dt=2\int (1+cos\,2t)dt=2t+sin\,2t+C$$
$$=2t+2sin\,tcos\,t+C$$
$$=2arcsin(\frac{x+1}{2})+\frac{x+1}{2}\sqrt{3-2x-x^{2}}+C$$
