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

To evaluate the integral $\int \sqrt{3-2 x-x^{2}} d x$,we first complete the square for the quadratic expression inside the square root.
$3-2x-x^2 = 4 - (x^2 + 2x + 1) = 4 - (x+1)^2$.
Thus,the integral becomes $\int \sqrt{4-(x+1)^2} d x$.
Let $x+1 = y$,then $dx = dy$.
The integral transforms to $\int \sqrt{2^2 - y^2} dy$.
Using the standard formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + C$,we get:
$= \frac{y}{2} \sqrt{4-y^2} + \frac{4}{2} \sin^{-1}(\frac{y}{2}) + C$.
Substituting $y = x+1$ back into the expression:
$= \frac{x+1}{2} \sqrt{4-(x+1)^2} + 2 \sin^{-1}(\frac{x+1}{2}) + C$.
$= \frac{x+1}{2} \sqrt{3-2x-x^2} + 2 \sin^{-1}(\frac{x+1}{2}) + C$.