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

Answer

To evaluate the integral \( \int (1 + x) \sqrt{1 - x} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = 1 - x \). Then, we have:
\[ x = 1 - t \quad \text{and} \quad dx = -dt \]

### Step 2: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral, we get:
\[ \int (1 + (1 - t)) \sqrt{t} (-dt) = \int (2 - t) \sqrt{t} (-dt) \]
This simplifies to:
\[ -\int (2 - t) \sqrt{t} \, dt \]

### Step 3: Distribute the Integral
Distributing the integral gives us:
\[ -\int (2\sqrt{t} - t\sqrt{t}) \, dt = -\int (2t^{1/2} - t^{3/2}) \, dt \]

### Step 4: Integrate Each Term
Now we can integrate each term separately:
\[ -\left( \int 2t^{1/2} \, dt - \int t^{3/2} \, dt \right) \]
Using the power rule for integration:
\[ \int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C \]
we find:
\[ -\left( 2 \cdot \frac{t^{3/2}}{3/2} - \frac{t^{5/2}}{5/2} \right) = -\left( \frac{4}{3} t^{3/2} - \frac{2}{5} t^{5/2} \right) \]

### Step 5: Simplify the Result
This simplifies to:
\[ -\frac{4}{3} t^{3/2} + \frac{2}{5} t^{5/2} + C \]

### Step 6: Substitute Back
Now, substitute \( t = 1 - x \) back into the expression:
\[ -\frac{4}{3} (1 - x)^{3/2} + \frac{2}{5} (1 - x)^{5/2} + C \]

### Final Result
Thus, the final answer is:
\[ \int (1 + x) \sqrt{1 - x} \, dx = -\frac{4}{3} (1 - x)^{3/2} + \frac{2}{5} (1 - x)^{5/2} + C \]