`int(x+2)sqrt(x^(2)+x)dx`

Answer

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

### Step 1: Rewrite the Integral
Let \( I = \int (x + 2) \sqrt{x^2 + x} \, dx \).

### Step 2: Express \( x + 2 \) in terms of the derivative of \( x^2 + x \)
Notice that \( \frac{d}{dx}(x^2 + x) = 2x + 1 \). We can express \( x + 2 \) as:
\[ x + 2 = \frac{3}{2}(2x + 1) - \frac{1}{2} \]
This means we can rewrite the integral as:
\[ I = \int \left( \frac{3}{2}(2x + 1) - \frac{1}{2} \right) \sqrt{x^2 + x} \, dx \]

### Step 3: Split the Integral
Now, we can split the integral into two parts:
\[ I = \frac{3}{2} \int (2x + 1) \sqrt{x^2 + x} \, dx - \frac{1}{2} \int \sqrt{x^2 + x} \, dx \]
Let \( I_1 = \int (2x + 1) \sqrt{x^2 + x} \, dx \) and \( I_2 = \int \sqrt{x^2 + x} \, dx \).

### Step 4: Solve \( I_2 \)
For \( I_2 \), we can use substitution. Let \( t = x^2 + x \), then \( dt = (2x + 1) \, dx \). Thus:
\[ I_2 = \int \sqrt{t} \cdot \frac{dt}{2} = \frac{1}{2} \int t^{1/2} \, dt = \frac{1}{2} \cdot \frac{2}{3} t^{3/2} + C = \frac{1}{3} (x^2 + x)^{3/2} + C \]

### Step 5: Solve \( I_1 \)
Now for \( I_1 \):
\[ I_1 = \int (2x + 1) \sqrt{x^2 + x} \, dx = \int \sqrt{x^2 + x} \, dt = \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} + C = \frac{2}{3} (x^2 + x)^{3/2} + C \]

### Step 6: Combine Results
Now substituting back into our expression for \( I \):
\[ I = \frac{3}{2} I_1 - \frac{1}{2} I_2 \]
Substituting the values of \( I_1 \) and \( I_2 \):
\[ I = \frac{3}{2} \cdot \frac{2}{3} (x^2 + x)^{3/2} - \frac{1}{2} \cdot \frac{1}{3} (x^2 + x)^{3/2} \]
This simplifies to:
\[ I = (x^2 + x)^{3/2} - \frac{1}{6} (x^2 + x)^{3/2} = \frac{5}{6} (x^2 + x)^{3/2} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int (x + 2) \sqrt{x^2 + x} \, dx = \frac{5}{6} (x^2 + x)^{3/2} + C \]