To integrate \(\frac{1}{1+x^2}\) from 0 to 1, we need to evaluate the definite integral:

\[
\int_{0}^{1} \frac{1}{1+x^2} \, dx
\]

The antiderivative of \(\frac{1}{1+x^2}\) is the arctangent function, \(\arctan(x)\). Therefore, the integral can be written as:

\[
\left. \arctan(x) \right|_0^1 = \arctan(1) - \arctan(0)
\]

Now, we evaluate the arctangent at the bounds:

\[
\arctan(1) = \frac{\pi}{4}, \quad \arctan(0) = 0
\]

So, the result of the integral is:

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

Thus, the value of the integral is \(\boxed{\frac{\pi}{4}}\).

Would you like a more detailed explanation or have any questions?

Here are some related questions:

1. What is the derivative of \(\arctan(x)\)?
2. How do you find the integral of \(\frac{1}{1-x^2}\)?
3. Can you evaluate \(\int_{0}^{\frac{\pi}{4}} \tan(x) \, dx\)?
4. What is the relationship between arctan and the unit circle?
5. How do you compute the integral of \(\frac{1}{x^2 + 4}\)?

**Tip:** Understanding the basic antiderivatives, like that of \(\frac{1}{1+x^2}\), helps in quickly solving integrals in calculus.

Explore the step-by-step solution for integrating the function 1/(1+x^2) over the interval [0,1]. This calculus problem demonstrates how to apply the arctan function and the Fundamental Theorem of Calculus. Perfect for students in Grades 11-12.

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