Evaluate the integrals : `I = int_(0)^(pi//4) (x^(2))/( x^(2) + 1) dx`,

Answer

To evaluate the integral

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

we can start by rewriting the integrand. We can express \( \frac{x^2}{x^2 + 1} \) as follows:

\[ \frac{x^2}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}. \]

This allows us to split the integral into two parts:

\[ I = \int_{0}^{\frac{\pi}{4}} \left( 1 - \frac{1}{x^2 + 1} \right) \, dx. \]

Now we can separate the integral:

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

### Step 1: Evaluate the first integral

The first integral is straightforward:

\[ \int_{0}^{\frac{\pi}{4}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

### Step 2: Evaluate the second integral

The second integral can be evaluated using the formula for the integral of \( \frac{1}{x^2 + a^2} \):

\[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) + C. \]

Thus,

\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{x^2 + 1} \, dx = \left[ \tan^{-1}(x) \right]_{0}^{\frac{\pi}{4}} = \tan^{-1}\left(\frac{\pi}{4}\right) - \tan^{-1}(0). \]

Since \( \tan^{-1}(0) = 0 \) and \( \tan^{-1}(1) = \frac{\pi}{4} \):

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

### Step 3: Combine the results

Now we can combine the results of the two integrals:

\[ I = \frac{\pi}{4} - \frac{\pi}{4} = 0. \]

Thus, the final result is:

\[ I = 0. \]