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 can break it down into two parts:

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

Let's denote the first integral as \( I_1 \) and the second as \( I_2 \):

1. **Evaluate \( I_1 = \int \frac{x}{\sqrt{3 - 2x - x^2}} \, dx \)**:

To simplify the integral, we can rewrite the expression under the square root. Notice that:

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

Thus, we have:

\[ I_1 = \int \frac{x}{\sqrt{-(x + 3)(x - 1)}} \, dx. \]

Now, we can multiply and divide by \(-2\):

\[ I_1 = -\frac{1}{2} \int \frac{-2x}{\sqrt{-(x + 3)(x - 1)}} \, dx. \]

We can rewrite this as:

\[ I_1 = -\frac{1}{2} \int \frac{-2x}{\sqrt{-(x + 3)(x - 1)}} \, dx. \]

Now, we can use substitution. Let \( t = 3 - 2x - x^2 \), then:

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

Substitute \( t \) into the integral to get:

\[ I_1 = -\frac{1}{2} \int \frac{-2x}{\sqrt{t}} \cdot \frac{dt}{-2(1 + x)}. \]

This simplifies to:

\[ I_1 = \frac{1}{2} \int \frac{x}{\sqrt{t}} \, dt. \]

2. **Evaluate \( I_2 = \int \frac{3}{\sqrt{3 - 2x - x^2}} \, dx \)**:

This integral can be simplified similarly:

\[ I_2 = 3 \int \frac{1}{\sqrt{3 - 2x - x^2}} \, dx. \]

Using the same substitution \( t = 3 - 2x - x^2 \), we can evaluate this integral as well.

The integral \( I_2 \) can be evaluated using trigonometric substitution or recognizing it as a standard integral form.

3. **Combine the results**:

After evaluating both \( I_1 \) and \( I_2 \), we can combine them to find the final result for \( I \).

### Final Answer:

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

where \( C \) is the constant of integration.