`int_(-2)^2(dx)/(1+|x-1|)` is equal to

Options

A. log2
B. 2 log2
C. 3 log2
D. 4 log2

Answer

To solve the integral \( \int_{-2}^{2} \frac{dx}{1 + |x - 1|} \), we will break it down into two parts based on the definition of the absolute value function.

### Step 1: Identify the points where the expression inside the absolute value changes
The expression \( |x - 1| \) changes at \( x = 1 \). Therefore, we will split the integral at this point.

### Step 2: Split the integral
We can write the integral as:
\[ \int_{-2}^{2} \frac{dx}{1 + |x - 1|} = \int_{-2}^{1} \frac{dx}{1 + |x - 1|} + \int_{1}^{2} \frac{dx}{1 + |x - 1|} \]

### Step 3: Evaluate the first integral \( \int_{-2}^{1} \frac{dx}{1 + |x - 1|} \)
For \( x < 1 \) (i.e., from \(-2\) to \(1\)), we have \( |x - 1| = 1 - x \). Therefore, the first integral becomes:
\[ \int_{-2}^{1} \frac{dx}{1 + (1 - x)} = \int_{-2}^{1} \frac{dx}{2 - x} \]

### Step 4: Evaluate the integral \( \int_{-2}^{1} \frac{dx}{2 - x} \)
The integral can be computed as:
\[ \int \frac{dx}{2 - x} = -\ln|2 - x| + C \]
Now, we evaluate it from \(-2\) to \(1\):
\[ \left[-\ln|2 - x|\right]_{-2}^{1} = -\ln|2 - 1| + \ln|2 - (-2)| = -\ln(1) + \ln(4) = 0 + \ln(4) = \ln(4) \]

### Step 5: Evaluate the second integral \( \int_{1}^{2} \frac{dx}{1 + |x - 1|} \)
For \( x \geq 1 \) (i.e., from \(1\) to \(2\)), we have \( |x - 1| = x - 1 \). Therefore, the second integral becomes:
\[ \int_{1}^{2} \frac{dx}{1 + (x - 1)} = \int_{1}^{2} \frac{dx}{x} \]

### Step 6: Evaluate the integral \( \int_{1}^{2} \frac{dx}{x} \)
The integral can be computed as:
\[ \int \frac{dx}{x} = \ln|x| + C \]
Now, we evaluate it from \(1\) to \(2\):
\[ \left[\ln|x|\right]_{1}^{2} = \ln(2) - \ln(1) = \ln(2) - 0 = \ln(2) \]

### Step 7: Combine the results
Now, we combine the results from 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:
\[ \int_{-2}^{2} \frac{dx}{1 + |x - 1|} = \ln(8) \]

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