If `intxlog(1+x^(2))dx=A(x).log(1+x^(2))+B(x)+c`, then

Options

A. `A(x)=((1+x^(2)))/(2)`
B. `B(x)=((1+x^(2)))/(2)`
C. `A(x)=-((1+x^(2)))/(2)`
D. `B(x)=-((1+x^(2)))/(2)`

Answer

To solve the integral \( I = \int x \log(1 + x^2) \, dx \) and express it in the form \( A(x) \log(1 + x^2) + B(x) + C \), we will use integration by parts. Here’s the step-by-step solution:

### Step 1: Set up the integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \log(1 + x^2) \) (which we will differentiate)
- \( dv = x \, dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now we differentiate \( u \) and integrate \( dv \):
- \( du = \frac{2x}{1 + x^2} \, dx \)
- \( v = \frac{x^2}{2} \)

### Step 3: Apply the integration by parts formula
Now we apply the integration by parts:
\[ I = \int x \log(1 + x^2) \, dx = \frac{x^2}{2} \log(1 + x^2) - \int \frac{x^2}{2} \cdot \frac{2x}{1 + x^2} \, dx \]
This simplifies to:
\[ I = \frac{x^2}{2} \log(1 + x^2) - \int \frac{x^3}{1 + x^2} \, dx \]

### Step 4: Simplify the integral
The integral \( \int \frac{x^3}{1 + x^2} \, dx \) can be simplified:
\[ \frac{x^3}{1 + x^2} = x - \frac{x}{1 + x^2} \]
Thus, we rewrite the integral:
\[ I = \frac{x^2}{2} \log(1 + x^2) - \int x \, dx + \int \frac{x}{1 + x^2} \, dx \]

### Step 5: Evaluate the remaining integrals
Now we evaluate the remaining integrals:
1. \( \int x \, dx = \frac{x^2}{2} \)
2. For \( \int \frac{x}{1 + x^2} \, dx \), we can use the substitution \( t = 1 + x^2 \), \( dt = 2x \, dx \):
\[ \int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \log(1 + x^2) \]

Putting it all together, we have:
\[ I = \frac{x^2}{2} \log(1 + x^2) - \frac{x^2}{2} + \frac{1}{2} \log(1 + x^2) + C \]
Combine the logarithmic terms:
\[ I = \left(\frac{x^2}{2} + \frac{1}{2}\right) \log(1 + x^2) - \frac{x^2}{2} + C \]

### Step 6: Identify \( A(x) \) and \( B(x) \)
From the expression, we can identify:
- \( A(x) = \frac{x^2 + 1}{2} \)
- \( B(x) = -\frac{x^2}{2} \)

### Final Result
Thus, we have:
\[ \int x \log(1 + x^2) \, dx = A(x) \log(1 + x^2) + B(x) + C \]
where \( A(x) = \frac{x^2 + 1}{2} \) and \( B(x) = -\frac{x^2}{2} \).