Evaluate the following integrals
Evaluate `int (2x-4)sqrt(4+3x-x^(2))dx`.

Answer

To evaluate the integral \( \int (2x - 4) \sqrt{4 + 3x - x^2} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = 4 + 3x - x^2 \). Now, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 3 - 2x \implies dt = (3 - 2x) \, dx \implies dx = \frac{dt}{3 - 2x} \]

### Step 2: Rewrite the Integral
Now we rewrite the integral in terms of \( t \):
\[ \int (2x - 4) \sqrt{t} \, dx = \int (2x - 4) \sqrt{t} \cdot \frac{dt}{3 - 2x} \]
To express \( 2x - 4 \) in terms of \( t \), we can rearrange \( t = 4 + 3x - x^2 \):
\[ 2x - 4 = 2x - 4 = - (3 - 2x) + 1 \]
Thus, we can substitute:
\[ \int (-(3 - 2x) + 1) \sqrt{t} \cdot \frac{dt}{3 - 2x} \]

### Step 3: Separate the Integral
This gives us two separate integrals:
\[ -\int \sqrt{t} \, dt + \int \frac{\sqrt{t}}{3 - 2x} \, dt \]

### Step 4: Evaluate the First Integral
The first integral can be evaluated as follows:
\[ -\int \sqrt{t} \, dt = -\frac{2}{3} t^{3/2} + C_1 \]

### Step 5: Evaluate the Second Integral
For the second integral, we need to express \( \frac{1}{3 - 2x} \) in terms of \( t \). From our substitution, we have:
\[ 3 - 2x = \sqrt{(4 + 3x - x^2)} \text{ (after some algebra)} \]
Thus, we can rewrite:
\[ \int \frac{\sqrt{t}}{3 - 2x} \, dt = \int \frac{\sqrt{t}}{\sqrt{t}} \, dt = \int dt = t + C_2 \]

### Step 6: Combine Results
Combining both results, we have:
\[ -\frac{2}{3} t^{3/2} + t + C \]

### Step 7: Substitute Back
Now substitute back \( t = 4 + 3x - x^2 \):
\[ -\frac{2}{3} (4 + 3x - x^2)^{3/2} + (4 + 3x - x^2) + C \]

### Final Result
The final result is:
\[ -\frac{2}{3} (4 + 3x - x^2)^{3/2} + (4 + 3x - x^2) + C \]