Consider `intxtan^(-1)xdx=A(x^(2)+1)tan^(-1)x+Bx+C` where , C is the constant of integration.
What is the value of A ?

Options

A. `1`
B. `(1)/(2)`
C. `-(1)/(2)`
D. `(1)/(4)`

Answer

To solve the integral \( \int x \tan^{-1}(x) \, dx \) and express it in the form \( A(x^2 + 1) \tan^{-1}(x) + Bx + C \), we will use integration by parts.

### Step-by-Step Solution:

1. **Identify the parts for integration by parts**:
- Let \( u = \tan^{-1}(x) \) and \( dv = x \, dx \).
- Then, \( du = \frac{1}{1+x^2} \, dx \) and \( v = \frac{x^2}{2} \).

2. **Apply integration by parts**:
\[ \int x \tan^{-1}(x) \, dx = u v - \int v \, du \]
Substituting the values we have:
\[ = \tan^{-1}(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx \]

3. **Simplify the integral**:
\[ = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx \]
The integral \( \int \frac{x^2}{1+x^2} \, dx \) can be simplified:
\[ \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} \]
Therefore,
\[ \int \frac{x^2}{1+x^2} \, dx = \int 1 \, dx - \int \frac{1}{1+x^2} \, dx = x - \tan^{-1}(x) \]

4. **Substitute back**:
\[ = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \left( x - \tan^{-1}(x) \right) \]
\[ = \frac{x^2}{2} \tan^{-1}(x) - \frac{x}{2} + \frac{1}{2} \tan^{-1}(x) \]

5. **Combine terms**:
\[ = \left( \frac{x^2}{2} + \frac{1}{2} \right) \tan^{-1}(x) - \frac{x}{2} + C \]
\[ = \frac{(x^2 + 1)}{2} \tan^{-1}(x) - \frac{x}{2} + C \]

6. **Compare with the given form**:
The expression we derived is:
\[ \frac{(x^2 + 1)}{2} \tan^{-1}(x) - \frac{x}{2} + C \]
This matches the form \( A(x^2 + 1) \tan^{-1}(x) + Bx + C \) where:
- \( A = \frac{1}{2} \)
- \( B = -\frac{1}{2} \)

### Conclusion:
The value of \( A \) is \( \frac{1}{2} \).