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

Options

A. `1-(pi)/4`
B. `(pi)/4-1`
C. `1-(pi)/2`
D. `0`

Answer

To solve the integral \( I = \int_0^1 \frac{x^2}{1+x^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We can rewrite the integrand by adding and subtracting 1 in the numerator:
\[ I = \int_0^1 \frac{x^2 + 1 - 1}{1+x^2} \, dx = \int_0^1 \left( \frac{x^2 + 1}{1+x^2} - \frac{1}{1+x^2} \right) \, dx \]
This simplifies to:
\[ I = \int_0^1 \left( 1 - \frac{1}{1+x^2} \right) \, dx \]

### 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 can be evaluated using the formula for the integral of \( \frac{1}{1+x^2} \):
\[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C \]
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 combine the results of the two integrals:
\[ I = 1 - \frac{\pi}{4} \]

### Final Result
Thus, the value of the integral is:
\[ I = 1 - \frac{\pi}{4} \]

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