How to solve $\int{\frac{1}{\sqrt{3-2x-x^2}}\,dx}$? $$\int{\frac{1}{\sqrt{3-2x-x^2}}\,dx}$$
I tried to do it by substitution with no sucess. Anyone can solve it?

Completing the square and using the substitution $u=x+1$ gives us 
\begin{equation*}
\int\frac{1}{\sqrt{4-(x+1)^2}}dx=\int\frac{1}{\sqrt{4-u^2}}du=\frac{1}{2}\int\frac{1}{\sqrt{1-\frac{u^2}{4}}}.
\end{equation*}
Using the substitution $s=\frac{u}{2}$ gives 
\begin{equation*}
\int\frac{1}{\sqrt{1-s^2}}=\sin^{-1}(s)+C.
\end{equation*}
Substituting back $s=\frac{u}{2}$ and $u=x+1$ gives the result. 
