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?
 A: $$\frac{x}{\sqrt{3-2x-x^2}} = -\frac{1}{2}\left(\frac{-2x}{\sqrt{3-2x-x^2}}\right)\\= -\frac{1}{2}\left(\frac{-2-2x+2}{\sqrt{3-2x-x^2}}\right)\\ =-\frac{1}{2}\left(\frac{-2-2x}{\sqrt{3-2x-x^2}}\right) - \frac{1}{\sqrt{3-2x-x^2}} \\= -\frac{1}{2}\left(\frac{-2-2x}{\sqrt{3-2x-x^2}}\right) - \frac{1}{\sqrt{4-(x+1)^2}}.$$  
Taking the integral, the first integrand will be an integral in the form
$$\frac{u'}{\sqrt{u}}$$
and the second is a standard integral that evaluates to a sine inverse.
A: Hint: You have done till here
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx  = \int \frac{x}{\sqrt{(4)-(x+1)^2}} dx$$
now let $x+1=2\sin t$ and simplify!
A: use the Eulersubstution and set $$\sqrt{3-2x-x^2}=xt\pm \sqrt{3}$$
A: $$\int \frac{x+1}{\sqrt{3-2x-x^2}}dx=-\frac{1}{2}\int\frac{-2-2x}{\sqrt{3-2x-x^2}}dx=-\frac{1}{2}\int\frac{g'(x)}{\sqrt{g(x)}}dx$$
$$=-\frac{1}{2}\int g'(x)[g(x)]^{-1/2}dx=-(3-2x-x^2)^{1/2}=-\sqrt{3-2x-x^2}+C$$
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 $g=3-2x-x^2$.
A: Integration strategy:
By adding/subtracting a constant to the numerator, you can let the derivative of the polynomial under the radical appear and you can integrate the ratio.
Then remains the inverse of the square root of this quadratic polynomial. By completing the square, you see that by translating by $1$ ($\to4-u^2$) and scaling the variable by $2$, you can reduce to $1-v^2$, giving an arc sine.
Hence the answer will be of the form
$$a\sqrt{3-2x-x^2}+b\arcsin\left(\frac{x+1}2\right).$$
By differentiating this expression (mentally) you can even guess that $a=-1$ to yield $x$ at the numerator. As this numerator will be $x+1$, the $+1$ is compensated with $b=-1$ (as $\arcsin'(u/2)=1/\sqrt{4-u^2}$ there is no extra factor).
