Evaluate the following integrals:
`int_(-2)^(2) (dx)/(1+|x-1|)`

Answer

To evaluate the integral

\[ \int_{-2}^{2} \frac{dx}{1 + |x - 1|}, \]

we first need to understand the behavior of the absolute value function \(|x - 1|\).

### Step 1: Analyze the absolute value function
The expression \(|x - 1|\) can be defined piecewise:
- For \(x < 1\), \(|x - 1| = -(x - 1) = -x + 1\).
- For \(x \geq 1\), \(|x - 1| = x - 1\).

### Step 2: Split the integral at the point where the absolute value changes
The integral from \(-2\) to \(2\) can be split into two parts:
\[ \int_{-2}^{2} \frac{dx}{1 + |x - 1|} = \int_{-2}^{1} \frac{dx}{1 + (-x + 1)} + \int_{1}^{2} \frac{dx}{1 + (x - 1)}. \]

### Step 3: Simplify each integral
1. For the first integral from \(-2\) to \(1\):
\[ \int_{-2}^{1} \frac{dx}{1 + (-x + 1)} = \int_{-2}^{1} \frac{dx}{2 - x}. \]

2. For the second integral from \(1\) to \(2\):
\[ \int_{1}^{2} \frac{dx}{1 + (x - 1)} = \int_{1}^{2} \frac{dx}{x}. \]

### Step 4: Evaluate the first integral
To evaluate \(\int_{-2}^{1} \frac{dx}{2 - x}\):
\[ \int \frac{dx}{2 - x} = -\ln|2 - x| + C. \]
Now, applying the limits:
\[ \left[-\ln|2 - x|\right]_{-2}^{1} = -\ln|2 - 1| + \ln|2 - (-2)| = -\ln(1) + \ln(4) = 0 + \ln(4) = \ln(4). \]

### Step 5: Evaluate the second integral
To evaluate \(\int_{1}^{2} \frac{dx}{x}\):
\[ \int \frac{dx}{x} = \ln|x| + C. \]
Now, applying the limits:
\[ \left[\ln|x|\right]_{1}^{2} = \ln(2) - \ln(1) = \ln(2) - 0 = \ln(2). \]

### Step 6: Combine the results
Now, we combine the results of both integrals:
\[ \int_{-2}^{2} \frac{dx}{1 + |x - 1|} = \ln(4) + \ln(2) = \ln(4 \cdot 2) = \ln(8). \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-2}^{2} \frac{dx}{1 + |x - 1|} = \ln(8). \]