Evaluate the following integrals.
` int (1 + x) sqrt(1-x)dx `

Answer

To evaluate the integral \( \int (1 + x) \sqrt{1 - x} \, dx \), we will use substitution and integration techniques.

### Step-by-Step Solution:

1. **Set up the integral:**
\[ I = \int (1 + x) \sqrt{1 - x} \, dx \]

2. **Use substitution:**
Let \( 1 - x = t^2 \). Then, differentiating both sides gives:
\[ -dx = 2t \, dt \quad \Rightarrow \quad dx = -2t \, dt \]

3. **Express \( x \) in terms of \( t \):**
From the substitution \( 1 - x = t^2 \), we can express \( x \) as:
\[ x = 1 - t^2 \]

4. **Substitute into the integral:**
Substitute \( x \) and \( dx \) into the integral:
\[ I = \int (1 + (1 - t^2)) \sqrt{t^2} (-2t) \, dt \]
Simplifying this gives:
\[ I = \int (2 - t^2) t (-2) \, dt = -2 \int (2t - t^3) \, dt \]

5. **Separate the integral:**
\[ I = -2 \left( \int 2t \, dt - \int t^3 \, dt \right) \]

6. **Integrate each term:**
- For \( \int 2t \, dt \):
\[ \int 2t \, dt = 2 \cdot \frac{t^2}{2} = t^2 \]
- For \( \int t^3 \, dt \):
\[ \int t^3 \, dt = \frac{t^4}{4} \]

7. **Combine the results:**
\[ I = -2 \left( t^2 - \frac{t^4}{4} \right) = -2t^2 + \frac{1}{2}t^4 \]

8. **Substitute back for \( t \):**
Recall that \( t = \sqrt{1 - x} \):
\[ t^2 = 1 - x \quad \text{and} \quad t^4 = (1 - x)^2 \]
Thus,
\[ I = -2(1 - x) + \frac{1}{2}(1 - x)^2 \]

9. **Simplify:**
\[ I = -2 + 2x + \frac{1}{2}(1 - 2x + x^2) = -2 + 2x + \frac{1}{2} - x + \frac{x^2}{2} \]
\[ I = \left(2x - x\right) + \left(-2 + \frac{1}{2}\right) + \frac{x^2}{2} = x + \frac{x^2}{2} - \frac{3}{2} \]

10. **Add the constant of integration:**
\[ I = x + \frac{x^2}{2} - \frac{3}{2} + C \]

### Final Answer:
\[ \int (1 + x) \sqrt{1 - x} \, dx = x + \frac{x^2}{2} - \frac{3}{2} + C \]