Can you explain the result in this one, please?

Question

I tried complete the square, but it doesn't work

I got this:

$\int\frac{xdx}{\sqrt{3-2x-x^2}}$

And I know that the answer is:

$-{\sqrt{3-2x-x^2}}-\arcsin(\frac{x+1}{2})+c$

Answer 1

As $$3-2x-x^2=2^2-(x+1)^2$$

start with $x+2=2\sin y$ where $-\dfrac\pi2\le y\le\dfrac\pi2$

$\implies dx=2\cos y\ dy$

$$\int\dfrac{x\ dx}{\sqrt{3-2x-x^2}}=\int\dfrac{2(\sin y-1)2\cos y}{+2\cos y}\ dy=?$$