The integral `int_(-1//2)^(1//2) ([x] + ln ((1+x)/(1-x)))dx` equals

Options

A. `-(1)/(2)`
B. 0
C. 1
D. `2ln ((1)/(2))`

Answer

To solve the integral

\[ I = \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( [x] + \ln \left( \frac{1+x}{1-x} \right) \right) dx, \]

we can separate it into two parts:

\[ I = I_1 + I_2, \]

where

\[ I_1 = \int_{-\frac{1}{2}}^{\frac{1}{2}} [x] \, dx \]

and

\[ I_2 = \int_{-\frac{1}{2}}^{\frac{1}{2}} \ln \left( \frac{1+x}{1-x} \right) \, dx. \]

### Step 1: Calculate \( I_2 \)

To evaluate \( I_2 \), we can use the property of definite integrals:

\[ \int_{-a}^{a} f(x) \, dx = 0 \quad \text{if } f(-x) = -f(x). \]

Let

\[ f(x) = \ln \left( \frac{1+x}{1-x} \right). \]

Now, we compute \( f(-x) \):

\[ f(-x) = \ln \left( \frac{1-x}{1+x} \right) = -\ln \left( \frac{1+x}{1-x} \right) = -f(x). \]

Since \( f(-x) = -f(x) \), we have:

\[ I_2 = \int_{-\frac{1}{2}}^{\frac{1}{2}} \ln \left( \frac{1+x}{1-x} \right) \, dx = 0. \]

### Step 2: Calculate \( I_1 \)

Now we calculate \( I_1 \):

\[ I_1 = \int_{-\frac{1}{2}}^{\frac{1}{2}} [x] \, dx. \]

The greatest integer function \( [x] \) is defined as follows in the interval \([-1/2, 1/2]\):

- For \( x \in \left[-\frac{1}{2}, 0\right) \), \( [x] = -1 \).
- For \( x \in \left[0, \frac{1}{2}\right) \), \( [x] = 0 \).

Thus, we can break \( I_1 \) into two parts:

\[ I_1 = \int_{-\frac{1}{2}}^{0} -1 \, dx + \int_{0}^{\frac{1}{2}} 0 \, dx. \]

Calculating each part:

1. For the first integral:

\[ \int_{-\frac{1}{2}}^{0} -1 \, dx = -1 \cdot \left(0 - \left(-\frac{1}{2}\right)\right) = -1 \cdot \frac{1}{2} = -\frac{1}{2}. \]

2. For the second integral:

\[ \int_{0}^{\frac{1}{2}} 0 \, dx = 0. \]

Putting it all together:

\[ I_1 = -\frac{1}{2} + 0 = -\frac{1}{2}. \]

### Final Calculation

Now we combine \( I_1 \) and \( I_2 \):

\[ I = I_1 + I_2 = -\frac{1}{2} + 0 = -\frac{1}{2}. \]

Thus, the value of the integral is

\[ \boxed{-\frac{1}{2}}. \]