`int_(0)^(1)x tan^(-1)x dx=`

Options

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

Answer

To solve the integral \( I = \int_{0}^{1} x \tan^{-1}(x) \, dx \), we will use the method of integration by parts.

### Step-by-Step Solution:

1. **Identify the parts for integration by parts**:
We will let:
- \( u = \tan^{-1}(x) \) (first function)
- \( dv = x \, dx \) (second function)

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

3. **Apply the integration by parts formula**:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
Applying this, we have:
\[ I = \left[ \tan^{-1}(x) \cdot \frac{x^2}{2} \right]_{0}^{1} - \int_{0}^{1} \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx \]

4. **Evaluate the boundary term**:
Now we evaluate the boundary term:
\[ \left[ \tan^{-1}(x) \cdot \frac{x^2}{2} \right]_{0}^{1} = \left( \tan^{-1}(1) \cdot \frac{1^2}{2} \right) - \left( \tan^{-1}(0) \cdot \frac{0^2}{2} \right) = \left( \frac{\pi}{4} \cdot \frac{1}{2} \right) - 0 = \frac{\pi}{8} \]

5. **Simplify the remaining integral**:
Now we need to simplify the remaining integral:
\[ \int_{0}^{1} \frac{x^2}{2(1+x^2)} \, dx \]
We can factor out \( \frac{1}{2} \):
\[ = \frac{1}{2} \int_{0}^{1} \frac{x^2}{1+x^2} \, dx \]

6. **Simplify the integrand**:
We can rewrite \( \frac{x^2}{1+x^2} \):
\[ \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} \]
Thus,
\[ \int_{0}^{1} \frac{x^2}{1+x^2} \, dx = \int_{0}^{1} 1 \, dx - \int_{0}^{1} \frac{1}{1+x^2} \, dx \]
The first integral evaluates to 1:
\[ = 1 - \left[ \tan^{-1}(x) \right]_{0}^{1} = 1 - \left( \frac{\pi}{4} - 0 \right) = 1 - \frac{\pi}{4} \]

7. **Combine results**:
Now substituting back:
\[ I = \frac{\pi}{8} - \frac{1}{2} \left( 1 - \frac{\pi}{4} \right) = \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} = \frac{2\pi}{8} - \frac{1}{2} = \frac{\pi}{4} - \frac{1}{2} \]

8. **Final result**:
Therefore, the final result is:
\[ I = \frac{\pi}{4} - \frac{1}{2} \]