The area bounded by the curves y=lnx, y=ln|x|, y=|lnx| and y=|ln||x| is

Answer

To find the area bounded by the curves \( y = \ln x \), \( y = \ln |x| \), \( y = |\ln x| \), and \( y = |\ln |x| \), we can follow these steps:

### Step 1: Understand the curves
1. **Curve \( y = \ln x \)**: This curve is defined for \( x > 0 \) and approaches negative infinity as \( x \) approaches 0 from the right.
2. **Curve \( y = \ln |x| \)**: This curve is defined for \( x \neq 0 \) and behaves like \( y = \ln x \) for \( x > 0 \) and like \( y = \ln (-x) \) for \( x < 0 \).
3. **Curve \( y = |\ln x| \)**: This curve is defined for \( x > 0 \) and is equal to \( \ln x \) for \( x > 1 \) and \( -\ln x \) for \( 0 < x < 1 \).
4. **Curve \( y = |\ln |x|| \)**: This curve behaves similarly to \( |\ln x| \) for \( x > 0 \) and \( |\ln (-x)| \) for \( x < 0 \).

### Step 2: Sketch the graphs
- Sketch the graphs of these functions to visualize the bounded areas. The curves will intersect at points where \( y = \ln x \) and \( y = |\ln x| \) meet, and similarly for the negative side.

### Step 3: Identify the bounded areas
- The bounded areas are symmetric about the y-axis. Therefore, we can calculate the area in one quadrant and multiply it by 4 to get the total area.

### Step 4: Set up the integral
- The area in the first quadrant (where \( x > 0 \)) can be calculated by integrating the difference between the curves \( |\ln x| \) and \( \ln x \) from \( x = 1 \) to \( x = e \) (where \( \ln x = 1 \)).
- The area in the second quadrant (where \( x < 0 \)) will be the same due to symmetry.

### Step 5: Calculate the area
1. **Area in the first quadrant**:
\[ A_1 = \int_1^e (|\ln x| - \ln x) \, dx = \int_1^e (0) \, dx = 0 \]
- For \( 0 < x < 1 \):
\[ A_2 = \int_0^1 (|\ln x| - \ln x) \, dx = \int_0^1 (-\ln x - \ln x) \, dx = \int_0^1 -2\ln x \, dx \]

2. **Evaluate the integral**:
- The integral of \( -\ln x \) can be computed using integration by parts:
\[ \int -\ln x \, dx = -x \ln x + x + C \]
- Evaluate from 0 to 1:
\[ A_2 = \left[-2(-x \ln x + x)\right]_0^1 = \left[-2(0 + 1) - \lim_{x \to 0} -2(-x \ln x + x)\right] \]
- As \( x \to 0 \), \( -x \ln x \to 0 \), thus:
\[ A_2 = 2 \]

3. **Total area**:
\[ \text{Total Area} = 4 \times A_2 = 4 \times 1 = 4 \]

### Final Result
The area bounded by the curves is \( 4 \).