If f(x)`=(x-1)/(x+1),f^(2)(x)=f(f(x)),……..,……..f^(k+1)(x)=f(f^(k)(x))`,k=1,2,3,……and `g(x)=f^(1998)(x)` then `int_(1//e)^(1) g(x)dx` is equal to

Options

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

Explanation

To solve the problem step by step, we will follow the definitions and calculations provided in the video transcript.

### Step 1: Define the function f(x)
Given:
\[ f(x) = \frac{x - 1}{x + 1} \]

### Step 2: Calculate f²(x)
To find \( f²(x) = f(f(x)) \):
1. Substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{x - 1}{x + 1}\right) \]
2. Calculate:
\[ f\left(\frac{x - 1}{x + 1}\right) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} \]
3. Simplify the numerator:
\[ \frac{x - 1 - (x + 1)}{x + 1} = \frac{x - 1 - x - 1}{x + 1} = \frac{-2}{x + 1} \]
4. Simplify the denominator:
\[ \frac{x - 1 + (x + 1)}{x + 1} = \frac{x - 1 + x + 1}{x + 1} = \frac{2x}{x + 1} \]
5. Thus, we have:
\[ f²(x) = \frac{-2/(x + 1)}{2x/(x + 1)} = \frac{-2}{2x} = \frac{-1}{x} \]

### Step 3: Calculate f⁴(x)
Next, we calculate \( f⁴(x) = f(f²(x)) \):
1. Substitute \( f²(x) \):
\[ f(f²(x)) = f\left(\frac{-1}{x}\right) \]
2. Calculate:
\[ f\left(\frac{-1}{x}\right) = \frac{\frac{-1}{x} - 1}{\frac{-1}{x} + 1} \]
3. Simplify the numerator:
\[ \frac{-1 - x}{x} = \frac{-(x + 1)}{x} \]
4. Simplify the denominator:
\[ \frac{-1 + x}{x} = \frac{x - 1}{x} \]
5. Thus:
\[ f⁴(x) = \frac{-(x + 1)/x}{(x - 1)/x} = \frac{-(x + 1)}{(x - 1)} = \frac{1 + x}{1 - x} \]

### Step 4: Calculate g(x) = f^(1998)(x)
Continuing this pattern, we can see that:
- \( f^2(x) = \frac{-1}{x} \)
- \( f^4(x) = x \)
- \( f^6(x) = \frac{-1}{x} \)
- \( f^8(x) = x \)

This indicates a periodicity of 2:
- If \( k \) is even, \( f^k(x) = x \)
- If \( k \) is odd, \( f^k(x) = \frac{-1}{x} \)

Since \( 1998 \) is even:
\[ g(x) = f^{1998}(x) = x \]

### Step 5: Evaluate the integral
Now we need to evaluate:
\[ \int_{1/e}^{1} g(x) \, dx = \int_{1/e}^{1} x \, dx \]
Calculating the integral:
\[ \int x \, dx = \frac{x^2}{2} \Big|_{1/e}^{1} = \frac{1^2}{2} - \frac{(1/e)^2}{2} = \frac{1}{2} - \frac{1/e^2}{2} = \frac{1}{2} - \frac{1}{2e^2} \]
Thus:
\[ = \frac{1}{2} - \frac{1}{2e^2} = \frac{e^2 - 1}{2e^2} \]

### Final Answer
The value of the integral is:
\[ \frac{e^2 - 1}{2e^2} \]