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 follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integrand:
\[ I = \int_{0}^{8} \frac{x^2}{x^2 + 8x + 32} \, dx \]
We can add and subtract \( 8x + 32 \) in the numerator:
\[ I = \int_{0}^{8} \left( \frac{x^2 + 8x + 32}{x^2 + 8x + 32} - \frac{8x + 32}{x^2 + 8x + 32} \right) \, dx \]
This simplifies to:
\[ I = \int_{0}^{8} 1 \, dx - \int_{0}^{8} \frac{8x + 32}{x^2 + 8x + 32} \, dx \]

### Step 2: Evaluate the First Integral
The first integral is straightforward:
\[ \int_{0}^{8} 1 \, dx = [x]_{0}^{8} = 8 - 0 = 8 \]

### Step 3: Simplify the Second Integral
Now we focus on the second integral:
\[ \int_{0}^{8} \frac{8x + 32}{x^2 + 8x + 32} \, dx \]
We can factor out the 8:
\[ = 8 \int_{0}^{8} \frac{x + 4}{x^2 + 8x + 32} \, dx \]

### Step 4: Substitution
Let \( t = x^2 + 8x + 32 \). Then, differentiating gives:
\[ dt = (2x + 8) \, dx \implies dx = \frac{dt}{2x + 8} \]
We also need to change the limits. When \( x = 0 \), \( t = 32 \) and when \( x = 8 \), \( t = 160 \).

### Step 5: Change of Variables
Now we substitute:
\[ \int_{0}^{8} \frac{x + 4}{x^2 + 8x + 32} \, dx = \int_{32}^{160} \frac{\frac{t - 32}{2} + 4}{t} \cdot \frac{dt}{2x + 8} \]
This requires solving for \( x \) in terms of \( t \). However, we can also directly evaluate:
\[ = \int_{32}^{160} \frac{dt}{t} \]

### Step 6: Evaluate the Logarithm Integral
The integral \( \int \frac{dt}{t} \) gives:
\[ \int_{32}^{160} \frac{dt}{t} = \ln(160) - \ln(32) = \ln\left(\frac{160}{32}\right) = \ln(5) \]

### Step 7: Combine Results
Now we can combine the results:
\[ I = 8 - 8 \cdot \ln(5) \]

### Step 8: Final Calculation
Using the approximate value \( \ln(5) \approx 1.609 \):
\[ I \approx 8 - 8 \cdot 1.609 = 8 - 12.872 = -4.872 \]

### Final Result
Thus, the value of the integral is:
\[ I = 8 - 8 \ln(5) \]