The value of the integral `int_(0)^(8)(x^(2))/(x^(2)+8x+32) dx` is equal to

Answer

To solve the integral \( I = \int_{0}^{8} \frac{x^2}{x^2 + 8x + 32} \, dx \), we can simplify the expression inside the integral.

### Step-by-step solution:

1. **Rewrite the integrand**:
We can rewrite the integrand by adding and subtracting \( 8x + 32 \) in the numerator:
\[ I = \int_{0}^{8} \left( \frac{x^2 + 8x + 32 - (8x + 32)}{x^2 + 8x + 32} \right) \, dx \]
This simplifies to:
\[ I = \int_{0}^{8} \left( 1 - \frac{8x + 32}{x^2 + 8x + 32} \right) \, dx \]

2. **Separate the integral**:
Now we can separate the integral into two parts:
\[ I = \int_{0}^{8} 1 \, dx - \int_{0}^{8} \frac{8x + 32}{x^2 + 8x + 32} \, dx \]

3. **Evaluate the first integral**:
The first integral is straightforward:
\[ \int_{0}^{8} 1 \, dx = [x]_{0}^{8} = 8 - 0 = 8 \]

4. **Evaluate the second integral**:
For the second integral, we can use substitution. Let:
\[ u = x^2 + 8x + 32 \quad \Rightarrow \quad du = (2x + 8) \, dx \quad \Rightarrow \quad dx = \frac{du}{2x + 8} \]
We also need to change the limits:
- When \( x = 0 \), \( u = 0^2 + 8 \cdot 0 + 32 = 32 \)
- When \( x = 8 \), \( u = 8^2 + 8 \cdot 8 + 32 = 128 \)

The integral becomes:
\[ \int_{32}^{128} \frac{8x + 32}{u} \cdot \frac{du}{2x + 8} \]
Notice that \( 8x + 32 = 4(2x + 8) \), so:
\[ \int_{32}^{128} \frac{4(2x + 8)}{u} \cdot \frac{du}{2x + 8} = 4 \int_{32}^{128} \frac{1}{u} \, du \]

5. **Compute the integral**:
Now we can compute:
\[ 4 \int_{32}^{128} \frac{1}{u} \, du = 4 [\ln |u|]_{32}^{128} = 4 (\ln 128 - \ln 32) = 4 \ln \left( \frac{128}{32} \right) = 4 \ln 4 \]

6. **Combine the results**:
Now we can combine the results:
\[ I = 8 - 4 \ln 4 \]

7. **Final simplification**:
Since \( \ln 4 = \ln(2^2) = 2 \ln 2 \):
\[ I = 8 - 8 \ln 2 \]

### Final Answer:
Thus, the value of the integral is:
\[ \int_{0}^{8} \frac{x^2}{x^2 + 8x + 32} \, dx = 8 - 8 \ln 2 \]