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?
 A: HINT: Complete the square $$3 - 2x - x^2 = 4-1-2x-x^2 =2^2 - (x+1)^2$$
Then put $(x+1) = 2\sin \theta$.
A: The thing you should think of instantly when you see a thing like that is that completing the square is the standard method in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial with no first-degree term.
In this case $3-2x-x^2 = 4 - (x+1)^2 = 4-u^2$.  Now you have a constant term, $4$, and a quadratic term, $-u^2$.
Next: Do you have
\begin{align}
\sqrt{a^2-u^2} \\[6pt]
\text{or }\sqrt{u^2-a^2} \\[6pt]
\text{or }\sqrt{u^2+a^2} & {}\quad \text{?}
\end{align}
Each yields to a different trigonometric subsitution.
A: 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. 
