`int_(-1)^(1)log((1+x)/(1-x))dx=`

Options

A. 2
B. 1
C. 0
D. -1

Answer

To solve the integral \( I = \int_{-1}^{1} \log\left(\frac{1+x}{1-x}\right) \, dx \), we can use the property of definite integrals that states:

\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} \left( f(x) + f(-x) \right) \, dx \]

### Step 1: Define the function
Let \( f(x) = \log\left(\frac{1+x}{1-x}\right) \).

### Step 2: Find \( f(-x) \)
Now, we need to compute \( f(-x) \):

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

### Step 3: Combine \( f(x) \) and \( f(-x) \)
Next, we add \( f(x) \) and \( f(-x) \):

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

Using the property of logarithms \( \log a + \log b = \log(ab) \):

\[ f(x) + f(-x) = \log\left(\frac{1+x}{1-x} \cdot \frac{1-x}{1+x}\right) = \log(1) = 0 \]

### Step 4: Set up the integral
Now we can set up the integral:

\[ I = \int_{-1}^{1} f(x) \, dx = \int_{0}^{1} \left( f(x) + f(-x) \right) \, dx = \int_{0}^{1} 0 \, dx \]

### Step 5: Evaluate the integral
Since the integral of zero over any interval is zero:

\[ I = 0 \]

### Conclusion
Thus, the value of the integral is:

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

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