If `f(x)=(1)/(1-x)`, then `int(f(f(f(x))))dx=`

Options

A. `x+c`
B. `(x^(2))/(2)+c`
C. `(x^(3))/(3)+c`
D. `(x^(4))/(4)+c`

Answer

To solve the problem, we need to find the integral of \( f(f(f(x))) \) where \( f(x) = \frac{1}{1-x} \).

### Step 1: Find \( f(f(x)) \)

Given:
\[ f(x) = \frac{1}{1-x} \]

Now, we need to find \( f(f(x)) \):
\[ f(f(x)) = f\left(\frac{1}{1-x}\right) = \frac{1}{1 - \frac{1}{1-x}} \]

To simplify this, we find a common denominator:
\[ 1 - \frac{1}{1-x} = \frac{(1-x) - 1}{1-x} = \frac{-x}{1-x} \]

Thus,
\[ f(f(x)) = \frac{1}{\frac{-x}{1-x}} = -\frac{1-x}{x} = \frac{x-1}{x} \]

### Step 2: Find \( f(f(f(x))) \)

Now, we compute \( f(f(f(x))) \):
\[ f(f(f(x))) = f\left(\frac{x-1}{x}\right) = \frac{1}{1 - \frac{x-1}{x}} \]

Again, simplifying:
\[ 1 - \frac{x-1}{x} = \frac{x - (x-1)}{x} = \frac{1}{x} \]

Thus,
\[ f(f(f(x))) = \frac{1}{\frac{1}{x}} = x \]

### Step 3: Integrate \( f(f(f(x))) \)

Now we need to integrate \( f(f(f(x))) \):
\[ \int f(f(f(x))) \, dx = \int x \, dx \]

Using the power rule for integration:
\[ \int x \, dx = \frac{x^2}{2} + C \]

### Final Answer

Thus, the final answer is:
\[ \int f(f(f(x))) \, dx = \frac{x^2}{2} + C \]