To integrate the function $\frac{1}{1+x^2}$ over the interval $[0, 1]$, we'll proceed as follows:

Step 1: Recognize the Integral Formula

The function $\frac{1}{1+x^2}$ is a standard integrand, and its antiderivative is:

$\int \frac{1}{1+x^2} \, dx = \arctan(x) + C$

where $C$ is the constant of integration.

Step 2: Apply the Definite Integral

We need to evaluate the definite integral from 0 to 1:

$\int_0^1 \frac{1}{1+x^2} \, dx = \left[\arctan(x)\right]_0^1$

Step 3: Compute the Antiderivative at the Limits

Now, substitute the limits of integration into the antiderivative:

$\arctan(1) - \arctan(0)$

• $\arctan(1) = \frac{\pi}{4}$
• $\arctan(0) = 0$

Step 4: Final Answer

The result of the definite integral is:

$\frac{\pi}{4} - 0 = \frac{\pi}{4}$

Thus, the integral of $\frac{1}{1+x^2}$ from 0 to 1 is $\frac{\pi}{4}$.

Would you like further details or explanations on any step?

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1. What is the geometric interpretation of the arctan function?
2. How does the integral change if the limits are from $[0, a]$ where $a > 1$?
3. Can the integral be computed using a numerical method instead of analytically?
4. What are the properties of the arctan function that make this integral solvable?
5. How does this integral relate to inverse trigonometric functions?

Tip: In calculus, recognizing standard integrals can greatly simplify solving definite integrals. Familiarize yourself with common antiderivatives like arctan(x).