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 B ?

Options

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

Answer

To solve the integral \( \int x \tan^{-1}(x) \, dx \) and find the value of \( B \) in the expression \( A(x^2 + 1) \tan^{-1}(x) + Bx + C \), 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, we differentiate and integrate:
\[ du = \frac{1}{1+x^2} \, dx \quad \text{and} \quad v = \frac{x^2}{2}. \]

2. **Apply the integration by parts formula**:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du. \]
Substituting our values:
\[ \int x \tan^{-1}(x) \, dx = \tan^{-1}(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx. \]

3. **Simplify the integral**:
The integral simplifies to:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx. \]
We can rewrite \( \frac{x^2}{1+x^2} \) as:
\[ \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}. \]
Thus, we have:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \left( \int 1 \, dx - \int \frac{1}{1+x^2} \, dx \right). \]

4. **Calculate the remaining integrals**:
The integrals can be computed as:
\[ \int 1 \, dx = x \quad \text{and} \quad \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x). \]
Therefore:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \left( x - \tan^{-1}(x) \right) + C. \]

5. **Combine the terms**:
Combining the terms gives:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2}x + \frac{1}{2} \tan^{-1}(x) + C. \]
This can be rearranged to:
\[ \int x \tan^{-1}(x) \, dx = \frac{1}{2}(x^2 + 1) \tan^{-1}(x) - \frac{1}{2}x + C. \]

6. **Compare with the given expression**:
The expression we have is:
\[ A(x^2 + 1) \tan^{-1}(x) + Bx + C. \]
From our result, we can see that:
\[ A = \frac{1}{2}, \quad B = -\frac{1}{2}, \quad \text{and the constant } C \text{ remains as is.} \]

### Conclusion:
Thus, the value of \( B \) is:
\[ \boxed{-\frac{1}{2}}. \]