If `int(x+2)/(2x^(2)+6x+5)dx`
`=P int(4x+6)/(2x^(2)+6x+5)dx+(1)/(2) int (dx)/(2x^(2)+6x+5')`
Then the value of P is

Options

A. `(1)/(3)`
B. `(1)/(2)`
C. `(1)/(4)`
D. 2

Explanation

To solve the problem, we need to find the value of \( P \) in the equation:

\[ \int \frac{x+2}{2x^2 + 6x + 5} \, dx = P \int \frac{4x+6}{2x^2 + 6x + 5} \, dx + \frac{1}{2} \int \frac{1}{2x^2 + 6x + 5} \, dx \]

### Step 1: Rewrite the Numerator
We can express the numerator \( x + 2 \) in terms of the derivative of the denominator \( 2x^2 + 6x + 5 \).

Let:
\[ f(x) = 2x^2 + 6x + 5 \]
Then,
\[ \frac{d}{dx} f(x) = 4x + 6 \]

We can write:
\[ x + 2 = A \cdot \frac{d}{dx}(2x^2 + 6x + 5) + B \]

### Step 2: Set Up the Equation
Now we differentiate:
\[ x + 2 = A(4x + 6) + B \]

### Step 3: Compare Coefficients
Now we can compare coefficients from both sides:
1. For \( x \): \( 4A = 1 \)
2. For the constant term: \( 6A + B = 2 \)

### Step 4: Solve for A and B
From \( 4A = 1 \):
\[ A = \frac{1}{4} \]

Substituting \( A \) into the second equation:
\[ 6 \left(\frac{1}{4}\right) + B = 2 \\ \frac{3}{2} + B = 2 \\ B = 2 - \frac{3}{2} = \frac{1}{2} \]

### Step 5: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \frac{x+2}{2x^2 + 6x + 5} \, dx = \int \left( \frac{1}{4} \cdot \frac{d}{dx}(2x^2 + 6x + 5) + \frac{1}{2} \right) \frac{1}{2x^2 + 6x + 5} \, dx \]

This can be split into two integrals:
\[ = \frac{1}{4} \int \frac{4x + 6}{2x^2 + 6x + 5} \, dx + \frac{1}{2} \int \frac{1}{2x^2 + 6x + 5} \, dx \]

### Step 6: Compare with Given Equation
Now we can compare this with the original equation:
\[ \int \frac{x+2}{2x^2 + 6x + 5} \, dx = P \int \frac{4x+6}{2x^2 + 6x + 5} \, dx + \frac{1}{2} \int \frac{1}{2x^2 + 6x + 5} \, dx \]

From this, we can see that:
\[ P = \frac{1}{4} \]

### Final Answer
Thus, the value of \( P \) is:
\[ \boxed{\frac{1}{4}} \]