The value of `int_(0)^(1) (xdx)/( (x^(2) +1)^(2) )` is

Options

A. `1/2`
B. `1`
C. `1/3`
D. `1/4`

Answer

To solve the integral \( I = \int_{0}^{1} \frac{x \, dx}{(x^2 + 1)^2} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_{0}^{1} \frac{x \, dx}{(x^2 + 1)^2} \]

### Step 2: Use Substitution
We will use the substitution \( t = x^2 + 1 \). Then, we find the differential \( dt \):
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]

### Step 3: Change the Limits of Integration
When \( x = 0 \):
\[ t = 0^2 + 1 = 1 \]
When \( x = 1 \):
\[ t = 1^2 + 1 = 2 \]
So, the limits change from \( x = 0 \) to \( x = 1 \) into \( t = 1 \) to \( t = 2 \).

### Step 4: Substitute in the Integral
Now we substitute \( x \) and \( dx \) in the integral:
\[ I = \int_{1}^{2} \frac{x}{t^2} \cdot \frac{dt}{2x} \]
The \( x \) cancels out:
\[ I = \int_{1}^{2} \frac{1}{2t^2} \, dt \]

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

### Step 6: Evaluate the Limits
Now we evaluate the limits:
\[ I = \frac{1}{2} \left( -\frac{1}{2} + 1 \right) = \frac{1}{2} \left( \frac{1}{2} \right) = \frac{1}{4} \]

### Final Answer
Therefore, the value of the integral is:
\[ \boxed{\frac{1}{4}} \]