Evaluate the following integrals `int (x-3)/sqrt(3-2x-x^(2))dx`

Answer

To evaluate the integral

\[ I = \int \frac{x-3}{\sqrt{3-2x-x^2}} \, dx, \]

we will follow a systematic approach.

### Step 1: Simplifying the Integral
First, we can rewrite the expression under the square root:

\[ 3 - 2x - x^2 = -(x^2 + 2x - 3) = -(x^2 + 2x + 1 - 4) = -((x+1)^2 - 4) = 4 - (x+1)^2. \]

Thus, we can express the integral as:

\[ I = \int \frac{x-3}{\sqrt{4 - (x+1)^2}} \, dx. \]

### Step 2: Substitution
Next, we will use the substitution:

\[ t = 4 - (x+1)^2. \]

Calculating the differential \(dt\):

\[ dt = -2(x+1) \, dx \implies dx = \frac{dt}{-2(x+1)}. \]

We also need to express \(x\) in terms of \(t\):

\[ (x+1)^2 = 4 - t \implies x + 1 = \sqrt{4 - t} \implies x = \sqrt{4 - t} - 1. \]

### Step 3: Substituting into the Integral
Now substitute \(x\) and \(dx\) into the integral:

\[ I = \int \frac{(\sqrt{4 - t} - 1) - 3}{\sqrt{t}} \cdot \frac{dt}{-2\sqrt{4 - t}}. \]

This simplifies to:

\[ I = -\frac{1}{2} \int \frac{\sqrt{4 - t} - 4}{\sqrt{t} \sqrt{4 - t}} \, dt. \]

### Step 4: Splitting the Integral
We can split the integral into two parts:

\[ I = -\frac{1}{2} \left( \int \frac{1}{\sqrt{t}} \, dt - 4 \int \frac{1}{\sqrt{4 - t}} \, dt \right). \]

### Step 5: Evaluating the Integrals
1. The first integral:

\[ \int \frac{1}{\sqrt{t}} \, dt = 2\sqrt{t}. \]

2. The second integral can be evaluated using the substitution \(u = 4 - t\):

\[ \int \frac{1}{\sqrt{4 - t}} \, dt = 2\sqrt{4 - t}. \]

### Step 6: Combining the Results
Putting it all together, we have:

\[ I = -\frac{1}{2} \left( 2\sqrt{t} - 8\sqrt{4 - t} \right) + C. \]

### Step 7: Back Substitution
Now substitute back \(t = 3 - 2x - x^2\):

\[ I = -\frac{1}{2} \left( 2\sqrt{3 - 2x - x^2} - 8\sqrt{4 - (3 - 2x - x^2 + 1)^2} \right) + C. \]

### Final Answer
Thus, the final result is:

\[ I = -\sqrt{3 - 2x - x^2} + 4\sin^{-1}\left(\frac{x + 1}{2}\right) + C. \]