`int_(-4)^(4) log ((7-x )/(7+x)) dx=?`

Options

A. `4`
B. `-4`
C. `0`
D. `1`

Answer

To solve the integral \( I = \int_{-4}^{4} \log\left(\frac{7-x}{7+x}\right) \, dx \), we can follow these steps:

### Step 1: Define the Integral
Let
\[ I = \int_{-4}^{4} \log\left(\frac{7-x}{7+x}\right) \, dx \]

### Step 2: Check for Evenness or Oddness
To determine if the function is even or odd, we need to evaluate \( f(-x) \):
\[ f(x) = \log\left(\frac{7-x}{7+x}\right) \]
Now, substitute \( -x \) into the function:
\[ f(-x) = \log\left(\frac{7-(-x)}{7+(-x)}\right) = \log\left(\frac{7+x}{7-x}\right) \]

### Step 3: Simplify \( f(-x) \)
Using the property of logarithms, we can rewrite \( f(-x) \):
\[ f(-x) = \log\left(\frac{7+x}{7-x}\right) = -\log\left(\frac{7-x}{7+x}\right) = -f(x) \]
This shows that \( f(-x) = -f(x) \), which means that \( f(x) \) is an odd function.

### Step 4: Apply the Property of Odd Functions
Since \( f(x) \) is an odd function, we can use the property of integrals of odd functions:
\[ \int_{-a}^{a} f(x) \, dx = 0 \]
for any odd function \( f(x) \).

### Step 5: Conclude the Result
Therefore, we have:
\[ I = \int_{-4}^{4} \log\left(\frac{7-x}{7+x}\right) \, dx = 0 \]

### Final Answer
\[ \int_{-4}^{4} \log\left(\frac{7-x}{7+x}\right) \, dx = 0 \]

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