The value of `int_(0)^(2)((x^(2)-2x+4)sin(x-1))/(2x^(2)-4x+5)dx` is equal to

Answer

To solve the integral

\[ I = \int_{0}^{2} \frac{(x^2 - 2x + 4) \sin(x - 1)}{2x^2 - 4x + 5} \, dx, \]

we will follow a series of steps to simplify and evaluate it.

### Step 1: Rewrite the Integral
We can rewrite the integral as:

\[ I = \int_{0}^{2} \frac{(x^2 - 2x + 4) \sin(x - 1)}{(2x^2 - 4x + 5)} \, dx. \]

### Step 2: Completing the Square
Next, we complete the square for the quadratic expressions in the numerator and denominator.

1. **Numerator**:
\[ x^2 - 2x + 4 = (x - 1)^2 + 3. \]

2. **Denominator**:
\[ 2x^2 - 4x + 5 = 2(x^2 - 2x + \frac{5}{2}) = 2((x - 1)^2 + \frac{3}{2}). \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{2} \frac{((x - 1)^2 + 3) \sin(x - 1)}{2((x - 1)^2 + \frac{3}{2})} \, dx. \]

### Step 3: Substitution
Let \( t = x - 1 \). Then \( dx = dt \) and the limits change as follows:
- When \( x = 0 \), \( t = -1 \).
- When \( x = 2 \), \( t = 1 \).

The integral now becomes:

\[ I = \int_{-1}^{1} \frac{(t^2 + 3) \sin(t)}{2(t^2 + \frac{3}{2})} \, dt. \]

### Step 4: Simplifying the Integral
This can be simplified to:

\[ I = \frac{1}{2} \int_{-1}^{1} \frac{(t^2 + 3) \sin(t)}{(t^2 + \frac{3}{2})} \, dt. \]

### Step 5: Analyzing the Function
Let

\[ f(t) = \frac{(t^2 + 3) \sin(t)}{(t^2 + \frac{3}{2})}. \]

We need to check if \( f(t) \) is an odd function.

Calculating \( f(-t) \):

\[ f(-t) = \frac{((-t)^2 + 3) \sin(-t)}{((-t)^2 + \frac{3}{2})} = \frac{(t^2 + 3)(-\sin(t))}{(t^2 + \frac{3}{2})} = -f(t). \]

Since \( f(-t) = -f(t) \), \( f(t) \) is an odd function.

### Step 6: Evaluating the Integral
The integral of an odd function over a symmetric interval around zero is zero:

\[ I = \frac{1}{2} \int_{-1}^{1} f(t) \, dt = 0. \]

### Final Answer
Thus, the value of the integral is:

\[ \boxed{0}. \]