Finding $\int\frac{x}{\sqrt{3-2x-x^2}} dx$. I was looking for the integral of 

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

My work:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx  = \int \frac{x}{\sqrt{(4)-(x+1)^2}} dx $$
Then I often remember this integral $\frac{u}{\sqrt{a^2 - x^2}} du$. So I modified the above integral to 
look like the integral $\frac{u}{\sqrt{a^2 - x^2}}$.
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx  = \int \frac{x+1-1}{\sqrt{(4)-(x+1)^2}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx + \int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$$
I recognized the the last integral $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$ has the form $\int \frac{1}{\sqrt{a^2-u^2}} du$, where $a =2$ and $u = x+1$.
It's corresponding integral would be $\arcsin \left( \frac{u}{a}\right) + c$. 
Evaluating $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$, it would be $-\arcsin \left( \frac{x+1}{2}\right)$
Here's the problem: I couldn't find the integral of $\int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx, $ because my Table of Integral doesn't show what is the 
integral of $\frac{u}{\sqrt{a^2 - x^2}} du$.
How to evaluate the integral of $\frac{x}{\sqrt{3-2x-x^2}} dx$ properly?

$$\int \frac{x+1}{\sqrt{3-2x-x^2}}dx=-\frac{1}{2}\int\frac{-2-2x}{\sqrt{3-2x-x^2}}dx=-\frac{1}{2}\int\frac{g'(x)}{\sqrt{g(x)}}dx$$
$$=-\frac{1}{2}\int g'(x)[g(x)]^{-1/2}dx=-(3-2x-x^2)^{1/2}=-\sqrt{3-2x-x^2}+C$$
This is the chain rule:
$$\int g'(x)f(g(x))dx=F(g(x))+C$$
where $F'(x)=f(x)$. In this case $f=x^{-1/2}$ and $g=3-2x-x^2$.
