The value of `int("ln"|x|)/(xsqrt(1+ln|x|)`dx equals:

Options

A. `2/3sqrt((1+ln|x|)("ln|x|-2))+C`
B. `2/3sqrt((1+"ln"|x|)(ln|x|+2))+C`
C. `1/2sqrt((1+ln|x|)(ln|x|-2))+C`
D. `2sqrt(1+ln|x|(3ln|x|=2)+C`

Answer

To solve the integral \(\int \frac{\ln |x|}{x \sqrt{1 + \ln |x|}} \, dx\), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = 1 + \ln |x| \). Then, we differentiate both sides:
\[ dt = \frac{1}{x} \, dx \quad \Rightarrow \quad dx = x \, dt \]
Since \( t = 1 + \ln |x| \), we can express \(\ln |x|\) in terms of \(t\):
\[ \ln |x| = t - 1 \]

### Step 2: Express \(x\) in terms of \(t\)
From the substitution \(t = 1 + \ln |x|\), we can express \(x\) as:
\[ \ln |x| = t - 1 \quad \Rightarrow \quad |x| = e^{t-1} \quad \Rightarrow \quad x = e^{t-1} \text{ (considering } x > 0\text{)} \]

### Step 3: Rewrite the integral
Now we can rewrite the integral:
\[ \int \frac{\ln |x|}{x \sqrt{1 + \ln |x|}} \, dx = \int \frac{t - 1}{e^{t-1} \sqrt{t}} \cdot e^{t-1} \, dt \]
This simplifies to:
\[ \int \frac{t - 1}{\sqrt{t}} \, dt \]

### Step 4: Split the integral
Now we can split the integral:
\[ \int \frac{t - 1}{\sqrt{t}} \, dt = \int \left( \frac{t}{\sqrt{t}} - \frac{1}{\sqrt{t}} \right) dt = \int \left( \sqrt{t} - t^{-1/2} \right) dt \]

### Step 5: Integrate term by term
Now we integrate each term:
1. \(\int \sqrt{t} \, dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2}\)
2. \(\int t^{-1/2} \, dt = 2 t^{1/2}\)

Putting it all together:
\[ \int \left( \sqrt{t} - t^{-1/2} \right) dt = \frac{2}{3} t^{3/2} - 2 t^{1/2} + C \]

### Step 6: Substitute back for \(t\)
Now we substitute back \(t = 1 + \ln |x|\):
\[ = \frac{2}{3} (1 + \ln |x|)^{3/2} - 2 (1 + \ln |x|)^{1/2} + C \]

### Final Answer
Thus, the value of the integral is:
\[ \int \frac{\ln |x|}{x \sqrt{1 + \ln |x|}} \, dx = \frac{2}{3} (1 + \ln |x|)^{3/2} - 2 (1 + \ln |x|)^{1/2} + C \]