Compute the integrals :
`int_(0)^(1) x "arc tan x dx`

Answer

To compute the integral \( \int_{0}^{1} x \, \text{arctan}(x) \, dx \), we will use the method of integration by parts.

### Step 1: Identify \( u \) and \( dv \)
We will set:
- \( u = \text{arctan}(x) \) (which we will differentiate)
- \( dv = x \, dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now we compute \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{1+x^2} \, dx \]
- Integrate \( dv \):
\[ v = \int x \, dx = \frac{x^2}{2} \]

### Step 3: Apply the integration by parts formula
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values into the formula:
\[ \int_{0}^{1} x \, \text{arctan}(x) \, dx = \left[ \text{arctan}(x) \cdot \frac{x^2}{2} \right]_{0}^{1} - \int_{0}^{1} \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx \]

### Step 4: Evaluate the boundary term
Now we evaluate the boundary term:
\[ \left[ \text{arctan}(x) \cdot \frac{x^2}{2} \right]_{0}^{1} = \left( \text{arctan}(1) \cdot \frac{1^2}{2} \right) - \left( \text{arctan}(0) \cdot \frac{0^2}{2} \right) \]
Calculating this:
\[ = \left( \frac{\pi}{4} \cdot \frac{1}{2} \right) - \left( 0 \cdot 0 \right) = \frac{\pi}{8} \]

### Step 5: Simplify the remaining integral
Now we simplify the remaining integral:
\[ \int_{0}^{1} \frac{x^2}{2(1+x^2)} \, dx = \frac{1}{2} \int_{0}^{1} \frac{x^2}{1+x^2} \, dx \]
We can rewrite \( \frac{x^2}{1+x^2} \) as \( 1 - \frac{1}{1+x^2} \):
\[ \int_{0}^{1} \frac{x^2}{1+x^2} \, dx = \int_{0}^{1} 1 \, dx - \int_{0}^{1} \frac{1}{1+x^2} \, dx \]
Calculating these integrals:
- The first integral:
\[ \int_{0}^{1} 1 \, dx = 1 \]
- The second integral:
\[ \int_{0}^{1} \frac{1}{1+x^2} \, dx = \text{arctan}(x) \bigg|_{0}^{1} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

Thus,
\[ \int_{0}^{1} \frac{x^2}{1+x^2} \, dx = 1 - \frac{\pi}{4} \]

### Step 6: Combine everything
Now we substitute back into our expression:
\[ \int_{0}^{1} x \, \text{arctan}(x) \, dx = \frac{\pi}{8} - \frac{1}{2} \left( 1 - \frac{\pi}{4} \right) \]
This simplifies to:
\[ = \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} = \frac{\pi}{4} - \frac{1}{2} \]

### Final Answer
Thus, the final result is:
\[ \int_{0}^{1} x \, \text{arctan}(x) \, dx = \frac{\pi}{4} - \frac{1}{2} \]