`int_0^1tan^(-1)xdx=`

Options

A. `pi//4`
B. `(pi//4) +log 2`
C. `(pi//4)-1/2log2`
D. `(pi//2) +log2`

Answer

To solve the integral \( \int_0^1 \tan^{-1}(x) \, dx \), we will use the method of integration by parts.

### Step-by-Step Solution:

1. **Choose \( u \) and \( dv \)**:
Let:
\[ u = \tan^{-1}(x) \quad \text{and} \quad dv = dx \]

2. **Differentiate \( u \) and integrate \( dv \)**:
Then, we find:
\[ du = \frac{1}{1+x^2} \, dx \quad \text{and} \quad v = x \]

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

4. **Simplify the second integral**:
The integral \( \int x \cdot \frac{1}{1+x^2} \, dx \) can be simplified by using the substitution \( w = 1 + x^2 \), which gives \( dw = 2x \, dx \) or \( dx = \frac{dw}{2x} \). Thus:
\[ \int x \cdot \frac{1}{1+x^2} \, dx = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln(1+x^2) + C \]

5. **Combine the results**:
Now substituting back, we have:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C \]

6. **Evaluate the definite integral from 0 to 1**:
Now we need to evaluate:
\[ \int_0^1 \tan^{-1}(x) \, dx = \left[ x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) \right]_0^1 \]

- **Upper limit (1)**:
\[ 1 \cdot \tan^{-1}(1) - \frac{1}{2} \ln(1+1^2) = 1 \cdot \frac{\pi}{4} - \frac{1}{2} \ln(2) \]
This simplifies to:
\[ \frac{\pi}{4} - \frac{1}{2} \ln(2) \]

- **Lower limit (0)**:
\[ 0 \cdot \tan^{-1}(0) - \frac{1}{2} \ln(1+0^2) = 0 - \frac{1}{2} \ln(1) = 0 \]

7. **Final result**:
Therefore, the value of the definite integral is:
\[ \int_0^1 \tan^{-1}(x) \, dx = \left( \frac{\pi}{4} - \frac{1}{2} \ln(2) \right) - 0 = \frac{\pi}{4} - \frac{1}{2} \ln(2) \]

### Final Answer:
\[ \int_0^1 \tan^{-1}(x) \, dx = \frac{\pi}{4} - \frac{1}{2} \ln(2) \]