`int(1)/(x(logx))dx=?`

Options

A. `log|x|+C`
B. `(-2)/(x^(2))+C`
C. `(logx)^(2)+C`
D. `log|logx|+C`

Explanation

To solve the integral \( \int \frac{1}{x \log x} \, dx \), we will use a substitution method. Here are the steps:

### Step 1: Substitution
Let \( t = \log x \).

### Step 2: Differentiate
Now, differentiate \( t \) with respect to \( x \):
\[ dt = \frac{1}{x} \, dx \quad \Rightarrow \quad dx = x \, dt \]
Since \( x = e^t \) (from \( t = \log x \)), we can substitute \( dx \):
\[ dx = e^t \, dt \]

### Step 3: Substitute in the Integral
Now substitute \( t \) and \( dx \) into the integral:
\[ \int \frac{1}{x \log x} \, dx = \int \frac{1}{e^t \cdot t} \cdot e^t \, dt \]
This simplifies to:
\[ \int \frac{1}{t} \, dt \]

### Step 4: Integrate
The integral of \( \frac{1}{t} \) is:
\[ \int \frac{1}{t} \, dt = \log |t| + C \]

### Step 5: Substitute Back
Now substitute back \( t = \log x \):
\[ \log |t| + C = \log |\log x| + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{1}{x \log x} \, dx = \log |\log x| + C \]

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