`Ifintxlog(1+1/x)dx=f(x)log(x+1)+g(x)x^2+A x+C ,` then `f(x)=1/2x^2` (b) `g(x)=logx` `A=1` (d) none of these

Options

A. `f(x)=(1)/(2)x^(2)`
B. `g(x)=log x`
C. `A=1`
D. none of these

Explanation

To solve the integral \( \int x \log(1 + \frac{1}{x}) \, dx \) and express it in the form \( f(x) \log(x + 1) + g(x)x^2 + Ax + C \), we will follow the steps outlined below.

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int x \log(1 + \frac{1}{x}) \, dx \]
We can rewrite \( \log(1 + \frac{1}{x}) \) using the property of logarithms:
\[ \log(1 + \frac{1}{x}) = \log\left(\frac{x + 1}{x}\right) = \log(x + 1) - \log(x) \]
Thus, the integral becomes:
\[ \int x \left( \log(x + 1) - \log(x) \right) \, dx \]
This can be separated into two integrals:
\[ \int x \log(x + 1) \, dx - \int x \log(x) \, dx \]

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

#### For \( \int x \log(x + 1) \, dx \):
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:
\[ \int x \log(x + 1) \, dx = \frac{x^2}{2} \log(x + 1) - \int \frac{x^2}{2} \cdot \frac{1}{x + 1} \, dx \]

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

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

### Step 3: Combine Results
Now we combine the results from both integrals:
\[ \int x \log(1 + \frac{1}{x}) \, dx = \left( \frac{x^2}{2} \log(x + 1) - \int \frac{x^2}{2} \cdot \frac{1}{x + 1} \, dx \right) - \left( \frac{x^2}{2} \log(x) - \frac{x^2}{4} \right) \]

### Step 4: Simplify the Expression
After simplifying, we can express the integral in the required form:
\[ \int x \log(1 + \frac{1}{x}) \, dx = f(x) \log(x + 1) + g(x)x^2 + Ax + C \]
From the integration, we can identify:
- \( f(x) = \frac{1}{2} x^2 \)
- \( g(x) = \log(x) \)
- \( A = 1 \)

### Final Answer
Thus, we conclude that:
- \( f(x) = \frac{1}{2} x^2 \)
- \( g(x) = \log(x) \)
- \( A = 1 \)