The value of the integral `int_(0)^(4)(x^(2))/(x^(2)-4x+8)dx` is equal to

Answer

To solve the integral

\[ I = \int_{0}^{4} \frac{x^2}{x^2 - 4x + 8} \, dx, \]

we will use the property of definite integrals, which states that

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

### Step 1: Apply the property of definite integrals

Here, \( a = 0 \) and \( b = 4 \). Thus, \( a + b = 4 \), and we can rewrite the integral as follows:

\[ I = \int_{0}^{4} \frac{(4 - x)^2}{(4 - x)^2 - 4(4 - x) + 8} \, dx. \]

### Step 2: Simplify the expression

Now, we simplify \( (4 - x)^2 \) and the denominator:

1. **Numerator**:
\[ (4 - x)^2 = 16 - 8x + x^2. \]

2. **Denominator**:
\[ (4 - x)^2 - 4(4 - x) + 8 = (16 - 8x + x^2) - (16 - 4x) + 8 = x^2 - 4x + 8. \]

Thus, we have:

\[ I = \int_{0}^{4} \frac{16 - 8x + x^2}{x^2 - 4x + 8} \, dx. \]

### Step 3: Combine the two integrals

Now, we can add the original integral and the transformed integral:

\[ 2I = \int_{0}^{4} \left( \frac{x^2}{x^2 - 4x + 8} + \frac{16 - 8x + x^2}{x^2 - 4x + 8} \right) \, dx. \]

### Step 4: Combine the numerators

Combining the numerators gives:

\[ 2I = \int_{0}^{4} \frac{2x^2 - 8x + 16}{x^2 - 4x + 8} \, dx. \]

### Step 5: Factor out the common terms

Factoring out a 2 from the numerator:

\[ 2I = 2 \int_{0}^{4} \frac{x^2 - 4x + 8}{x^2 - 4x + 8} \, dx = 2 \int_{0}^{4} 1 \, dx. \]

### Step 6: Evaluate the integral

Now, evaluate the integral:

\[ 2I = 2 \left[ x \right]_{0}^{4} = 2(4 - 0) = 8. \]

### Step 7: Solve for \( I \)

Thus, we have:

\[ I = \frac{8}{2} = 4. \]

### Final Answer

The value of the integral is

\[ \boxed{4}. \]