Answer
$\displaystyle\int{\sqrt{3-2x-x^2}dx}=\displaystyle\frac{1}{2}\left( x+1 \right) \left( \sqrt{3-2x-x^2} \right) +2\arcsin \left( \frac{x+1}{2} \right) +C$
Work Step by Step
$
I=\displaystyle\int{\sqrt{3-2x-x^2}dx}\\
I=\displaystyle\int{\sqrt{3+1-1-2x-x^2}dx}\\
I=\int{\sqrt{4-\left( x+1 \right) ^2}dx}$
$\displaystyle{\left[\begin{matrix}
x+1=2\sin \theta& \left( x+1 \right) ^2=4\sin ^2\theta\\
\frac{dx}{d\theta}=2\cos \theta& dx=2\cos \theta d\theta\\
\end{matrix}\right]}$
$I=\displaystyle\int{\sqrt{4-4\sin ^2\theta}\ 2\cos \theta d\theta}\\
I=\displaystyle\int{2\cos \theta \times2\cos \theta \ d\theta}\\
I=\displaystyle\int{4\cos ^2\theta \ d\theta}\\
I=\displaystyle4\int{\frac{1}{2}\left( \cos 2\theta +1 \right) \ d\theta}\\
I=\displaystyle2\int{\cos 2\theta +1\ d\theta}\\
I=\displaystyle2\left( \frac{1}{2}\sin 2\theta +\theta \right) +C\\
I=\displaystyle\sin 2\theta +2\theta +C\\
I=\displaystyle2\sin \theta \cos \theta +2\theta +C$
$\displaystyle{\left[\begin{matrix}
\sin \theta =\frac{x+1}{2}& \cos \theta =\frac{\sqrt{3-2x-x^2}}{2}\\
\end{matrix}\right]}$
$I=\displaystyle2\left( \frac{x+1}{2} \right) \left( \frac{\sqrt{3-2x-x^2}}{2} \right) +2\arcsin \left( \frac{x+1}{2} \right) +C\\
I=\displaystyle\frac{1}{2}\left( x+1 \right) \left( \sqrt{3-2x-x^2} \right) +2\arcsin \left( \frac{x+1}{2} \right) +C
$
