If `f_(n) = log log … log x` (log is repeated n times) then `int [x f_(1)(x) f_(2)(x) … f_(n)(x)]^(-1) dx` is equal to

Options

A. `f_(n+1) (x) + C`
B. `(f_(n+1)(x))/(n+1)+C`
C. `nf_(n) (x) + C`
D. none of these

Answer

To solve the integral

\[ I = \int \left[ x f_1(x) f_2(x) \ldots f_n(x) \right]^{-1} dx \]

where \( f_n(x) = \log(\log(\ldots(\log(x))\ldots)) \) (with \( n \) logs), we proceed as follows:

### Step 1: Rewrite the Integral
We can rewrite the integral as:

\[ I = \int \frac{1}{x f_1(x) f_2(x) \ldots f_n(x)} \, dx \]

### Step 2: Change of Variables
Let \( f_n(x) = t \). Then, we need to find \( \frac{dt}{dx} \).

### Step 3: Differentiate \( f_n(x) \)
Using the chain rule, we differentiate \( f_n(x) \):

\[ \frac{d}{dx} f_n(x) = \frac{1}{x \cdot f_{n-1}(x)} \cdot \frac{d}{dx} f_{n-1}(x) \]

Continuing this process, we find that:

\[ \frac{d}{dx} f_n(x) = \frac{1}{x \cdot f_{n-1}(x) \cdot f_{n-2}(x) \ldots f_1(x)} \]

### Step 4: Substitute into the Integral
Now, substituting \( dt \) into the integral, we get:

\[ I = \int \frac{1}{t} dt \]

### Step 5: Integrate
The integral of \( \frac{1}{t} \) is:

\[ I = \log |t| + C \]

### Step 6: Substitute Back
Substituting back for \( t = f_n(x) \):

\[ I = \log |f_n(x)| + C \]

Since \( f_n(x) = \log(\log(\ldots(\log(x))\ldots)) \) (with \( n \) logs), we can express this as:

\[ I = \log \left( \log(\log(\ldots(\log(x))\ldots)) \right) + C \]

### Step 7: Final Expression
Finally, we recognize that we have \( n \) logs in \( f_n(x) \), and thus:

\[ I = \log \left( f_{n+1}(x) \right) + C \]

where \( f_{n+1}(x) = \log(\log(\ldots(\log(x))\ldots)) \) (with \( n+1 \) logs).

### Conclusion
The final result is:

\[ I = f_{n+1}(x) + C \]