What is the integral for this expression? (I tried complete the square) But complete the square, doesn´t lead me to something coherent.
I got this:
$$\int\frac{x \, dx}{\sqrt{3-2x-x^2}}$$
 A: $$
\int\frac{x \, dx}{\sqrt{3-2x-x^2}}
$$
First let $u=3-2x-x^2$ so that $du = -2(1+x)\,dx$ and $\dfrac{du}{-2} = (1+x)\,dx$.  Then we can say
\begin{align}
\int\frac{x \, dx}{\sqrt{3-2x-x^2}} = \int \frac{(1+x) \, dx}{\sqrt{3-2x-x^2}} - \int \frac{dx}{\sqrt{3-2x-x^2}}.
\end{align}
For the first integral, use the substitution.
Then we must deal with the second integral.  Complete the square:
$$
3-2x-x^2 = -(x^2 + 2x + 1) + 3+1 = 4 - (x+1)^2.
$$
So we have
$$
\int \frac{dx}{\sqrt{4-(x+1)^2}} = \int \frac{dx/2}{\sqrt{1 - \left( \frac{x+1}2 \right)^2}} = \int \frac{\cos\theta\,d\theta}{\sqrt{1-\sin^2\theta}} = \int 1\,d\theta = \text{etc.} 
$$
After that we need to change $\theta$ to $\arcsin\dfrac{x+1}{2}$ and so on.
A: OK, here is a first step: 
Since $D(3-2x-x^2)=-2(x+1)$ it might be a good idea to write
$$
\frac{x}{\sqrt{3-2x-x^2}}=\frac{x+1}{\sqrt{3-2x-x^2}}-\frac{1}{\sqrt{3-2x-x^2}}.
$$
Maybe you can integrate that term now?
For the second expression, you have already completed the square in the comment. Now, you should recognize that you get an arcsin, since
$$
\int\frac{1}{\sqrt{1-t^2}}\,dt=\arcsin t.
$$
What substitution can you do to get your second expression into this form (with some constant in front, maybe)? I.e., you want
$$
\frac{1}{\sqrt{3-2x-x^2}}=\frac{A}{\sqrt{1-t^2}},
$$
for some $A$.
