`int " x tan"^(-1) " x dx "`

Answer

To solve the integral \( \int x \tan^{-1}(x) \, dx \), we will use the method of integration by parts.

### Step-by-Step Solution:

1. **Identify \( u \) and \( dv \)**:
- Let \( u = \tan^{-1}(x) \) and \( dv = x \, dx \).

2. **Differentiate \( u \) and Integrate \( dv \)**:
- Differentiate \( u \):
\[ du = \frac{1}{1+x^2} \, dx \]
- Integrate \( dv \):
\[ v = \frac{x^2}{2} \]

3. **Apply the Integration by Parts Formula**:
- The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
- Applying this, we have:
\[ \int x \tan^{-1}(x) \, dx = \tan^{-1}(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx \]

4. **Simplify the Integral**:
- The integral simplifies to:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx \]

5. **Further Simplify the Integral**:
- The term \( \frac{x^2}{1+x^2} \) can be rewritten as:
\[ \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} \]
- Thus, the integral becomes:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \left( \int 1 \, dx - \int \frac{1}{1+x^2} \, dx \right) \]

6. **Evaluate the Remaining Integrals**:
- The integrals are:
\[ \int 1 \, dx = x \]
\[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \]
- Therefore, we have:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \left( x - \tan^{-1}(x) \right) \]

7. **Combine the Results**:
- Combining everything gives:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{x}{2} + \frac{1}{2} \tan^{-1}(x) + C \]
- Rearranging terms:
\[ = \frac{x^2}{2} \tan^{-1}(x) + \frac{1}{2} \tan^{-1}(x) - \frac{x}{2} + C \]

### Final Answer:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2}{2} \tan^{-1}(x) - \frac{x}{2} + \frac{1}{2} \tan^{-1}(x) + C \]