If `f(x)=int_(0)^(1)(dt)/(1+|x-t|),x in R`. The value of `f'(1//2)` is equal to

Options

A. `1//2`
B. `0`
C. `1`
D. `2`

Explanation

To find the value of \( f'(1/2) \) where

\[ f(x) = \int_{0}^{1} \frac{dt}{1 + |x - t|}, \]

we will follow these steps:

### Step 1: Analyze the Absolute Value
The absolute value \( |x - t| \) can be expressed in two cases:
- When \( x \geq t \), then \( |x - t| = x - t \).
- When \( x < t \), then \( |x - t| = t - x \).

### Step 2: Split the Integral
We can split the integral into two parts based on the value of \( x \):
\[ f(x) = \int_{0}^{x} \frac{dt}{1 + (x - t)} + \int_{x}^{1} \frac{dt}{1 + (t - x)}. \]

### Step 3: Simplify Each Integral
1. For the first integral:
\[ \int_{0}^{x} \frac{dt}{1 + (x - t)} = \int_{0}^{x} \frac{dt}{1 + x - t} = \int_{0}^{x} \frac{dt}{(1 + x) - t}. \]
Let \( u = 1 + x - t \), then \( du = -dt \) and the limits change from \( t = 0 \) to \( t = x \) which corresponds to \( u = 1 + x \) to \( u = 1 + x - x = 1 \):
\[ = -\int_{1 + x}^{1} \frac{du}{u} = \ln(1 + x) - \ln(1 + x) = \ln\left(\frac{1 + x}{1}\right) = \ln(1 + x). \]

2. For the second integral:
\[ \int_{x}^{1} \frac{dt}{1 + (t - x)} = \int_{x}^{1} \frac{dt}{1 + t - x} = \int_{x}^{1} \frac{dt}{(1 - x) + t}. \]
Let \( v = (1 - x) + t \), then \( dv = dt \) and the limits change from \( t = x \) to \( t = 1 \) which corresponds to \( v = 1 - x + x = 1 \) to \( v = 1 - x + 1 = 2 - x \):
\[ = \int_{1}^{2 - x} \frac{dv}{v} = \ln(2 - x) - \ln(1) = \ln(2 - x). \]

### Step 4: Combine the Results
Thus, we can express \( f(x) \) as:
\[ f(x) = \ln(1 + x) + \ln(2 - x) = \ln((1 + x)(2 - x)). \]

### Step 5: Differentiate \( f(x) \)
Now, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \ln((1 + x)(2 - x)) = \frac{1}{(1 + x)(2 - x)} \cdot \left( (2 - x) - (1 + x) \right) = \frac{(2 - x) - (1 + x)}{(1 + x)(2 - x)}. \]
Simplifying the numerator:
\[ = \frac{2 - x - 1 - x}{(1 + x)(2 - x)} = \frac{1 - 2x}{(1 + x)(2 - x)}. \]

### Step 6: Evaluate \( f'(1/2) \)
Now, we substitute \( x = \frac{1}{2} \):
\[ f'\left(\frac{1}{2}\right) = \frac{1 - 2 \cdot \frac{1}{2}}{(1 + \frac{1}{2})(2 - \frac{1}{2})} = \frac{1 - 1}{(1.5)(1.5)} = \frac{0}{2.25} = 0. \]

### Final Answer
Thus, the value of \( f'(1/2) \) is:
\[ \boxed{0}. \]