If `int\ x ln(1+1/x)\ dx = p(x)\ ln(1+1/x)+1/2 x - 1/2 ln(1+x) + c,` c being arbitarary constant, then

Options

A. `p(x)=(1)/(2)x^(2)`
B. `p(x)=0`
C. `p(x)=1`
D. none of these

Answer

To solve the integral \( \int x \ln\left(1 + \frac{1}{x}\right) dx \), we will use integration by parts. Let's go through the steps in detail.

### Step 1: Set up integration by parts
We will use the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \ln\left(1 + \frac{1}{x}\right) \)
- \( dv = x \, dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{d}{dx} \ln\left(1 + \frac{1}{x}\right) \, dx = \frac{1}{1 + \frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right) \, dx = -\frac{1}{x^2 + x} \, dx \]
- Integrate \( dv \):
\[ v = \int x \, dx = \frac{x^2}{2} \]

### Step 3: Apply integration by parts
Now substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula:
\[ \int x \ln\left(1 + \frac{1}{x}\right) dx = \frac{x^2}{2} \ln\left(1 + \frac{1}{x}\right) - \int \frac{x^2}{2} \left(-\frac{1}{x^2 + x}\right) dx \]

### Step 4: Simplify the integral
The integral simplifies to:
\[ \int x \ln\left(1 + \frac{1}{x}\right) dx = \frac{x^2}{2} \ln\left(1 + \frac{1}{x}\right) + \frac{1}{2} \int \frac{x^2}{x^2 + x} dx \]
Now, simplify \( \frac{x^2}{x^2 + x} = 1 - \frac{x}{x^2 + x} \):
\[ \int x \ln\left(1 + \frac{1}{x}\right) dx = \frac{x^2}{2} \ln\left(1 + \frac{1}{x}\right) + \frac{1}{2} \left( \int 1 \, dx - \int \frac{x}{x^2 + x} dx \right) \]

### Step 5: Evaluate the remaining integrals
1. The first integral:
\[ \int 1 \, dx = x \]
2. The second integral can be simplified using partial fractions:
\[ \int \frac{x}{x^2 + x} dx = \int \frac{1}{x + 1} dx = \ln|x + 1| \]

### Step 6: Combine results
Putting it all together:
\[ \int x \ln\left(1 + \frac{1}{x}\right) dx = \frac{x^2}{2} \ln\left(1 + \frac{1}{x}\right) + \frac{1}{2} \left( x - \ln|x + 1| \right) + C \]

### Step 7: Final expression
Thus, we have:
\[ \int x \ln\left(1 + \frac{1}{x}\right) dx = \frac{x^2}{2} \ln\left(1 + \frac{1}{x}\right) + \frac{1}{2} x - \frac{1}{2} \ln(x + 1) + C \]

### Step 8: Identify \( p(x) \)
Comparing with the given expression:
\[ p(x) \ln\left(1 + \frac{1}{x}\right) + \frac{1}{2} x - \frac{1}{2} \ln(x + 1) + C \]
we find that:
\[ p(x) = \frac{x^2}{2} \]