If `intx log (1+(1)/(x))dx`
`=f(x).log_(e)(x+1)+g(x)log_(e)x+Lx+C` , then

Options

A. `f(x)=(x^(2))/(2)`
B. `g(x)=log_(e)x`
C. L = 1
D. `L=(1)/(2)`

Explanation

To solve the integral \( \int x \log\left(1 + \frac{1}{x}\right) dx \) and express it in the form \( f(x) \log(x + 1) + g(x) \log x + Lx + C \), we will proceed step by step.

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int x \log\left(1 + \frac{1}{x}\right) dx \]
We can simplify the logarithmic expression:
\[ \log\left(1 + \frac{1}{x}\right) = \log\left(\frac{x + 1}{x}\right) = \log(x + 1) - \log(x) \]
Thus, we can rewrite the integral as:
\[ \int x \left(\log(x + 1) - \log x\right) dx = \int x \log(x + 1) dx - \int x \log x dx \]

### Step 2: Apply Integration by Parts
We will apply integration by parts to both integrals. Recall the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]

#### For \( I_1 = \int x \log(x + 1) dx \):
Let \( u = \log(x + 1) \) and \( dv = x \, dx \).
Then, \( du = \frac{1}{x + 1} dx \) and \( v = \frac{x^2}{2} \).

Applying integration by parts:
\[ I_1 = \frac{x^2}{2} \log(x + 1) - \int \frac{x^2}{2} \cdot \frac{1}{x + 1} dx \]

Now, simplify the integral:
\[ \int \frac{x^2}{2(x + 1)} dx = \frac{1}{2} \int \left(x - 1 + \frac{1}{x + 1}\right) dx \]
This can be separated into:
\[ \frac{1}{2} \left(\int x \, dx - \int 1 \, dx + \int \frac{1}{x + 1} dx\right) \]

Calculating these integrals:
\[ \int x \, dx = \frac{x^2}{2}, \quad \int 1 \, dx = x, \quad \int \frac{1}{x + 1} dx = \log(x + 1) \]

Thus,
\[ \int \frac{x^2}{2(x + 1)} dx = \frac{1}{2} \left(\frac{x^2}{2} - x + \log(x + 1)\right) \]

Combining everything, we have:
\[ I_1 = \frac{x^2}{2} \log(x + 1) - \frac{1}{4} x^2 + \frac{1}{2} x - \frac{1}{2} \log(x + 1) \]

#### For \( I_2 = \int x \log x \, dx \):
Let \( u = \log x \) and \( dv = x \, dx \).
Then, \( du = \frac{1}{x} dx \) and \( v = \frac{x^2}{2} \).

Applying integration by parts:
\[ I_2 = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx = \frac{x^2}{2} \log x - \frac{x^2}{4} \]

### Step 3: Combine \( I_1 \) and \( I_2 \)
Now we combine the results:
\[ \int x \log\left(1 + \frac{1}{x}\right) dx = I_1 - I_2 \]
Substituting \( I_1 \) and \( I_2 \):
\[ = \left(\frac{x^2}{2} \log(x + 1) - \frac{1}{4} x^2 + \frac{1}{2} x - \frac{1}{2} \log(x + 1)\right) - \left(\frac{x^2}{2} \log x - \frac{x^2}{4}\right) \]

### Step 4: Simplify the Expression
Combining all terms:
\[ = \frac{x^2}{2} \log(x + 1) - \frac{x^2}{2} \log x + \frac{1}{2} x - \frac{1}{2} \log(x + 1) + \frac{1}{4} x^2 \]

Now, we can express this in the required form:
\[ = f(x) \log(x + 1) + g(x) \log x + Lx + C \]
Where:
- \( f(x) = \frac{x^2}{2} - \frac{1}{2} \)
- \( g(x) = -\frac{x^2}{2} \)
- \( L = \frac{1}{2} \)

### Final Result
Thus, the values are:
- \( f(x) = \frac{x^2 - 1}{2} \)
- \( g(x) = -\frac{x^2}{2} \)
- \( L = \frac{1}{2} \)