`int_(0)^(1) (x^(2))/(1+x^(2))dx` is equal to

Options

A. `pi/4-1`
B. `1-pi/2`
C. `pi/2-1`
D. `1-pi/4`

Explanation

To solve the integral \( I = \int_{0}^{1} \frac{x^2}{1+x^2} \, dx \), we can use a clever technique by rewriting the integrand.

### Step 1: Rewrite the integrand
We can express the integrand as:
\[ \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} \]
Thus, we can rewrite the integral as:
\[ I = \int_{0}^{1} \left( 1 - \frac{1}{1+x^2} \right) \, dx \]

### Step 2: Split the integral
Now we can split the integral into two parts:
\[ I = \int_{0}^{1} 1 \, dx - \int_{0}^{1} \frac{1}{1+x^2} \, dx \]

### Step 3: Evaluate the first integral
The first integral is straightforward:
\[ \int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 - 0 = 1 \]

### Step 4: Evaluate the second integral
The second integral is:
\[ \int_{0}^{1} \frac{1}{1+x^2} \, dx \]
This integral can be evaluated using the formula for the arctangent:
\[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \]
Thus,
\[ \int_{0}^{1} \frac{1}{1+x^2} \, dx = \left[ \tan^{-1}(x) \right]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Step 5: Combine the results
Now we can substitute back into our expression for \( I \):
\[ I = 1 - \frac{\pi}{4} \]

### Final Result
Thus, the value of the integral is:
\[ \int_{0}^{1} \frac{x^2}{1+x^2} \, dx = 1 - \frac{\pi}{4} \]