`int_(-1)^(2) (x) /((x^(2)+1)^(2))dx`

Answer

To solve the integral \( I = \int_{-1}^{2} \frac{x}{(x^2 + 1)^2} \, dx \), we will use a substitution method. Here’s a step-by-step breakdown of the solution:

### Step 1: Substitution
Let \( t = x^2 + 1 \). Then, differentiate both sides:
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]

### Step 2: Change the limits of integration
When \( x = -1 \):
\[ t = (-1)^2 + 1 = 2 \]
When \( x = 2 \):
\[ t = 2^2 + 1 = 5 \]
Thus, the limits change from \( x = -1 \) to \( x = 2 \) into \( t = 2 \) to \( t = 5 \).

### Step 3: Rewrite the integral
Substituting \( t \) and \( dx \) into the integral, we have:
\[ I = \int_{2}^{5} \frac{x}{t^2} \cdot \frac{dt}{2x} \]
The \( x \) terms cancel out:
\[ I = \int_{2}^{5} \frac{1}{2t^2} \, dt \]

### Step 4: Integrate
Now, we can integrate:
\[ I = \frac{1}{2} \int_{2}^{5} t^{-2} \, dt \]
The integral of \( t^{-2} \) is:
\[ \int t^{-2} \, dt = -\frac{1}{t} \]
Thus,
\[ I = \frac{1}{2} \left[-\frac{1}{t}\right]_{2}^{5} \]

### Step 5: Evaluate the definite integral
Now, we evaluate the limits:
\[ I = \frac{1}{2} \left(-\frac{1}{5} + \frac{1}{2}\right) \]
Calculating this gives:
\[ I = \frac{1}{2} \left(\frac{1}{2} - \frac{1}{5}\right) = \frac{1}{2} \left(\frac{5}{10} - \frac{2}{10}\right) = \frac{1}{2} \cdot \frac{3}{10} = \frac{3}{20} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{3}{20} \]