Find $\int \sqrt{3 - 2x - x^2} \, dx$

$\frac{x + 1}{2}\sqrt{3 - 2x - x^2} + 2 \sin^{-1}\left(\frac{x + 1}{2}\right) + C$

The correct answer is Option (3) → $\frac{x + 1}{2}\sqrt{3 - 2x - x^2} + 2 \sin^{-1}\left(\frac{x + 1}{2}\right) + C$

$\int \sqrt{3 - 2x - x^2} \, dx = \int \sqrt{4 - (x+1)^2} \, dx$

Put $x + 1 = y$, so that $dx = dy$:

Thus $\int \sqrt{3 - 2x - x^2} \, dx =\int \sqrt{4 - y^2} \, dy$

$= \frac{1}{2} y \sqrt{4 - y^2} + \frac{4}{2} \sin^{-1} \frac{y}{2} + C$

$= \frac{1}{2} (x+1) \sqrt{3 - 2x - x^2} + 2 \sin^{-1} \left( \frac{x+1}{2} \right) + C$