The antiderivative of `f (x) = log (log x ) + 1/(log x)^(2)` whose graph passes through (e,e) is

Options

A. `x (log (log x) + (log x)^(-1))`
B. `x (log (log x)- (log x)^(-1)) + e`
C. `x(log(logx)-(logx)^(-1)+2e`
D. none of these

Answer

To find the antiderivative of the function \( f(x) = \log(\log x) + \frac{1}{(\log x)^2} \) that passes through the point \( (e, e) \), we will follow these steps:

### Step 1: Set up the integral
We need to find the integral of \( f(x) \):
\[ I = \int f(x) \, dx = \int \left( \log(\log x) + \frac{1}{(\log x)^2} \right) dx \]

### Step 2: Split the integral
We can split the integral into two parts:
\[ I = \int \log(\log x) \, dx + \int \frac{1}{(\log x)^2} \, dx \]
Let \( I_1 = \int \log(\log x) \, dx \) and \( I_2 = \int \frac{1}{(\log x)^2} \, dx \).

### Step 3: Integrate \( I_1 \) using integration by parts
For \( I_1 \), we use integration by parts. Let:
- \( u = \log(\log x) \) then \( du = \frac{1}{\log x} \cdot \frac{1}{x} \, dx \)
- \( dv = dx \) then \( v = x \)

Applying integration by parts:
\[ I_1 = x \log(\log x) - \int x \cdot \frac{1}{\log x} \cdot \frac{1}{x} \, dx \]
This simplifies to:
\[ I_1 = x \log(\log x) - \int \frac{1}{\log x} \, dx \]

### Step 4: Integrate \( I_2 \)
Now we need to integrate \( I_2 \):
\[ I_2 = \int \frac{1}{(\log x)^2} \, dx \]
Using substitution, let \( v = \log x \) then \( dv = \frac{1}{x} \, dx \) or \( dx = e^v \, dv \):
\[ I_2 = \int \frac{1}{v^2} e^v \, dv \]
This integral can be solved using integration by parts again, but we will keep it as is for now.

### Step 5: Combine the results
Combining \( I_1 \) and \( I_2 \):
\[ I = x \log(\log x) - \int \frac{1}{\log x} \, dx + \int \frac{1}{(\log x)^2} \, dx + C \]

### Step 6: Evaluate the constants using the point (e, e)
Now we need to find the constant \( C \) such that the graph passes through the point \( (e, e) \):
\[ I(e) = e \log(\log e) - \int \frac{1}{\log e} \, dx + \int \frac{1}{(\log e)^2} \, dx + C \]
Since \( \log e = 1 \):
\[ I(e) = e \cdot 0 - \int 1 \, dx + \int 1 \, dx + C \]
This simplifies to:
\[ e = -e + e + C \implies C = 2e \]

### Final Result
Thus, the antiderivative is:
\[ I = x \log(\log x) - \frac{x}{\log x} + 2e \]