Question:

Note that $\int \sqrt{3 - 2x - x^2} dx = \int \sqrt{4 - (x+1)^2} dx$ <br> Put $x + 1 = y$ so that $dx = dy$ <br> Thus $\int \sqrt{3 - 2x - x^2} dx = \int \sqrt{4 - y^2} dy = \frac{1}{2}y\sqrt{4 - y^2} + \frac{4}{2}\sin^{-1} \frac{y}{2} + C$ <br> $= \frac{1}{2}(x+1)\sqrt{3 - 2x - x^2} + 2\sin^{-1} \left(\frac{x+1}{2}\right) + C$