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 follow these steps:

### Step 1: Simplify the Denominator
First, we need to rewrite the denominator \( x^2 - 4x + 8 \) as a perfect square. We can do this by completing the square.

\[ x^2 - 4x + 8 = (x^2 - 4x + 4) + 4 = (x - 2)^2 + 4 \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral using the new form of the denominator:

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

### Step 3: Change of Variables
Next, we will use the substitution \( t = x - 2 \). Thus, \( x = t + 2 \) and \( dx = dt \). We also need to change the limits of integration:

- When \( x = 0 \), \( t = 0 - 2 = -2 \)
- When \( x = 4 \), \( t = 4 - 2 = 2 \)

Now the integral becomes:

\[ I = \int_{-2}^{2} \frac{(t + 2)^2}{t^2 + 4} \, dt \]

### Step 4: Expand the Numerator
Now we expand the numerator:

\[ (t + 2)^2 = t^2 + 4t + 4 \]

So we can rewrite the integral as:

\[ I = \int_{-2}^{2} \frac{t^2 + 4t + 4}{t^2 + 4} \, dt \]

### Step 5: Split the Integral
We can split the integral into three parts:

\[ I = \int_{-2}^{2} \frac{t^2}{t^2 + 4} \, dt + \int_{-2}^{2} \frac{4t}{t^2 + 4} \, dt + \int_{-2}^{2} \frac{4}{t^2 + 4} \, dt \]

### Step 6: Evaluate Each Integral
1. **First Integral:**
\[ \int_{-2}^{2} \frac{t^2}{t^2 + 4} \, dt \]
This integral is even, so we can compute it from 0 to 2 and double the result.

2. **Second Integral:**
\[ \int_{-2}^{2} \frac{4t}{t^2 + 4} \, dt = 0 \]
This integral is odd, and thus evaluates to zero.

3. **Third Integral:**
\[ \int_{-2}^{2} \frac{4}{t^2 + 4} \, dt = 4 \int_{-2}^{2} \frac{1}{t^2 + 4} \, dt \]
This integral can be evaluated using the formula for the integral of \( \frac{1}{a^2 + x^2} \).

### Step 7: Final Calculation
The integral \( \int \frac{1}{4 + t^2} \, dt \) can be computed as:

\[ \int \frac{1}{4 + t^2} \, dt = \frac{1}{2} \tan^{-1} \left( \frac{t}{2} \right) \]

Evaluating from -2 to 2 gives:

\[ \left[ \frac{1}{2} \tan^{-1} \left( \frac{2}{2} \right) - \frac{1}{2} \tan^{-1} \left( \frac{-2}{2} \right) \right] = \frac{1}{2} \left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]

Thus, the third integral contributes \( 4 \cdot \frac{\pi}{4} = \pi \).

### Conclusion
Combining all parts, we find:

\[ I = 2 \cdot \left( \frac{\pi}{4} \right) + 0 + \pi = \frac{\pi}{2} + \pi = \frac{3\pi}{2} \]

Thus, the value of the integral is:

\[ \boxed{4} \]